Get Ready for Grade 8 - This area is password protected …petrinmath.weebly.com/.../pom10_solutions_ch2_final.pdfChapter 2 Analytic Geometry Chapter 2 Get Ready Chapter 2 Get Ready

Chapter 2 Analytic Geometry Chapter 2 Get Ready Chapter 2 Get Ready Question 1 Page 54 a) 3 2 14

3 2 2 14 23 123 123 3

4

xx

xx

x

+ =+ − = −

=

=

=

b) 7 5 2 10

7 5 2 5 2 10 25 155 155 5

3

y yy y y y

yy

y

− = +− − + = + − +

=

=

=

5

c)

1 1 24 31 112 12 12 24 33 4 24

3 4 4 24 42424

1 124

z z

z z

z zz z z z

zz

z

= −

× = × − ×

= −− = − −− = −− −

=− −

= d)

( ) ( )2 2

0.5

0.5

0.25

t

t

t

=

=

=

MHR • Principles of Mathematics 10 Solutions 1

Chapter 2 Get Ready Question 2 Page 54 a) 2 0

2 2 02

21 1

2

x yx y x x

y xy x

y x

− + =− + − − = − −

− = − −− −

= −− − −

= +

2

1

5

b) 3 5 0

3 5 3 5 0 33 5

x yx y x x

y x

+ − =+ − − + = − +

= − + c) 2 4 7 0

2 4 7 2 7 0 2 74 2 74 24 4

1 72 4

x yx y x x

y xy x

y x

− + =− + − − = − −

− = − −− −

= −− − −

= +

74

d)

1 3 5 02

1 1 13 5 5 0 52 2 2

13 521

3 523 3

1 56 3

x y

x y x x

y x

y x

y x

− + =

− + − − = − −

− = − −

−−= −

− − −

= +

3

2 MHR • Principles of Mathematics 10 Solutions

Chapter 2 Get Ready Question 3 Page 54

a) 12am =

b) 2412

bm =

=

c) 2323

cm −=

= −

d) 2814

am −=

= −



a) 2 1

2 1

10 612 44812

y ymx x−

=−−

=

=

=

−

b)

( )

2 1

2 1

461

1

4

y ym

2 621 4

x x−

=−

=

−=

= −

−− −

c)

( )( )

2 1

2 1

18

2

4

y ymx x−

=−

−

−

=

=

= −

8 43 5−− −−

d) 2 1

2 1

7.6 6.49.8 2.51.27.31273

y ymx x−

=−−

=−

=

=


Chapter 2 Get Ready Question 5 Page 55 a)

2 4−y mx b

x= += +

The equation of the line is y = –2x + 4. b)

2 147

y mx b

x

= +

= −

The equation of the line is y = 27

x – 14.

c)

( )

( )

3

1

= +4 63 24

21

4 24 2

1

y mx bb

bb

y mx bxx

= +

= +− =

−

= +

= +

= −The equation of the line is y = 4x + 21. d)

( )14 22

1 32

= +

−

−−

4 13

y mx b

b

bb

y mx b

x

= +

= +

+

=

=

=

+

The equation of the line is y = 12

− x + 3.



a) 2 1

2 1

9 15 1

48

2

y ymx x−

=−

=

=

=

2 11

y mx bb

b

= +

− =

−−

Substitute m = 2 and the coordinates of one point, say (1, 1), to find b.

1= +

( )

The equation of the line is y = 2x – 1. b)

( )

2 1

2 1

32

2 13 1

−−−

−

23

y ymx x−

=−

=−

−=−

=

Substitute m = 32

and the coordinates of one point, say (–1, 1), to find b.

( )31 12

= +−

31252

y mx b

b

b

b

= +

+ =

=

The equation of the line is 3 52 2

y x= + .


c)

( )

2 1

2 1

4 12 436

21

y ymx x−

=−

=−

=

=

−−

Substitute m = 12


( )112

= +4

3

y mx b

b

b

−

= +

=

The equation of the line is 1 32

y x= + .

d)

( )

2 1

2 1

12

26

y ym

4 81 5

x x−

=−

=−

=

= −

52

y mx bb

b

= +

=

−−−−

8 2= +−−

Substitute m = –2 and the coordinates of one point, say (5, –8), to find b.

( )

The equation of the line is y = –2x + 2.


Chapter 2 Get Ready Question 7 Page 55 a) The slope of the given line is 3. The slope of a line parallel to the given line is also 3.

b) The slope of the given line is 16

− . The slope of a line parallel to the given line is also 16

− .

c) The slope of the given line is –4. The slope of a line perpendicular to the given line is the

negative reciprocal, 14

.

d) The slope of the given line is 34

. The slope of a line perpendicular to the given line is the

negative reciprocal, 43

− .


Chapter 2 Get Ready Question 8 Page 55 a) The slope of the required line is –3.

( )5 = +34

3y mx b

bb− −

= +

− =

The equation of the line is y = –3x – 4.

b) The slope of the required line is 23

.

( )433

9 43

3 23

3

2

53

y mx b

b

b

b

b

= +

= +

= +

− =

=


y x= + .

c) The slope of the required line is 34

− .

( )1514

4 1

31 54

= +−−

54 4

14

1

y mx b

b

b

b

b

= +

= +

− =

− =


y x= − −1 .


Chapter 2 Get Ready Question 9 Page 55 a) The measure of is the same as the measure of

, or 60º. D∠

C∠

b)

35 6

EF EDAB ACEF

5 3EF6

EF 2.5

=

=

×=

=

The length of side EF is 2.5 cm. Chapter 2 Get Ready Question 10 Page 55 If P is any point on the right bisector of line segment AB and Q is the point of intersection of AB and the right bisector, then AQ = QB and

PQA = PQB = 90°. ∠ ∠ Side PQ is common to ΔPQA and ΔPQB. Therefore, ΔPQA is congruent to ΔPQB (side-angle-side). PA and PB are corresponding sides, so PA = PB.


Chapter 2 Section 1: Midpoint of a Line Segment Chapter 2 Section 1 Question 1 Page 66 a) The midpoint of AB is at (4, 6). b) The midpoint of CD is at (1, 3). c) The midpoint of EF is at (2, 2).


d) The midpoint of GH is at 1 , 22

⎛ ⎞− −⎜ ⎟⎝ ⎠

.


Chapter 2 Section 1 Question 2 Page 66

a) ( )

( )

1 2 1 2, ,2 2

,2 2

5 3

8 16,2 24

7 9

,8

x x y yx y + +⎛ ⎞= ⎜ ⎟⎝ ⎠

+ +⎛ ⎞= ⎜ ⎟⎝ ⎠⎛ ⎞= ⎜ ⎟⎝ ⎠

=

The coordinates of the midpoint of J(5, 7) and K(3, 9) are (4, 8).

b) ( )

( )

( )

1 2 1 2

0 6

, ,2 2

,2 2

0 6,

1

2 20, 3

1

x x y yx y + +⎛ ⎞= ⎜ ⎟⎝ ⎠⎛ ⎞++

= ⎜ ⎟⎜ ⎟⎝ ⎠

−⎛ ⎞= ⎜ ⎟⎝ ⎠

=

− −

− The coordinates of the midpoint of L(–1, 0) and M(1, –6) are (0, –3).

c) ( )

( )

( )

1 2 1 2, ,2 2

,2 2

4 4,2 2

2 2 4 8

2,2

x x y yx y

− − −

+ +⎛ ⎞= ⎜ ⎟⎝ ⎠⎛ ⎞+ +

= ⎜ ⎟⎜ ⎟⎝ ⎠−⎛ ⎞= ⎜ ⎟

⎝ ⎠= −

The coordinates of the midpoint of N(–2, –4) and P(–2, 8) are (–2, 2).

d) ( )

( ) ( )

( )

1 2 1 2, ,2 2

2 2

4 10,2 22, 5

x x y yx y + +⎛ ⎞= ⎜ ⎟⎝ ⎠⎛ ⎞

= ⎜ ⎟⎜ ⎟⎝ ⎠− −⎛ ⎞= ⎜ ⎟

⎝ ⎠= − −

,13 3 7+ +− − − −

The coordinates of the midpoint of Q(–3, –3) and R(–1, –7) are (–2, –5).



a) ( )

( )

1 2 1 2

0.2 3

, ,2 2

,.6 1.2 2

3.8 1.7,2 2

1.

5

9,0.85

0.2

x x y yx y + +⎛ ⎞= ⎜ ⎟⎝ ⎠

+ +⎛ ⎞= ⎜ ⎟⎝ ⎠⎛ ⎞= ⎜ ⎟⎝ ⎠

=

The coordinates of the midpoint of J(0.2, 1.5) and K(3.6, 0.2) are (1.9, 0.85).

b) ( )

( )

( )

1 2 1 2

3.2 5.31.4 0.6

, ,2 2

,2 2

0.8 8.5,2 2

0.4, 4.25

x x y yx y + +⎛ ⎞= ⎜ ⎟⎝ ⎠⎛ ⎞++

= ⎜ ⎟⎜ ⎟⎝ ⎠− −⎛ ⎞= ⎜ ⎟

⎝ ⎠

−−−

−−= The coordinates of the midpoint of N(–1.4, –3.2) and P(0.6, –5.3) are (–0.4, –4.25). c)

( )

( )

1 2 1 2

5 51 32 2

, ,2 2

,2 2

2 0,2 2

1,0

2 2

x x y yx y + +⎛ ⎞= ⎜ ⎟⎝ ⎠⎛ ⎞⎛ ⎞++⎜ ⎟⎜ ⎟

⎝ ⎠⎜ ⎟=⎜ ⎟⎜ ⎟⎝ ⎠⎛ ⎞= ⎜ ⎟⎝

=

−

⎠

The coordinates of the midpoint of L 1 5,2 2

⎛⎜⎝ ⎠

⎞⎟ and M 3 5,

2 2⎛ −⎜⎝ ⎠

⎞⎟ are (1, 0).


d)

( ) 1 2 1 2

1 73 2 8 88

, ,2 2

,2 2

13 68 8,2 2

13 3,16 8

x x y yx y + +⎛ ⎞= ⎜ ⎟⎝ ⎠⎛ ⎞⎛ ⎞++⎜ ⎟⎜ ⎟

⎝ ⎠⎜ ⎟=⎜ ⎟⎜ ⎟⎝ ⎠⎛ ⎞−⎜ ⎟

−−

= ⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎛ ⎞= −⎜⎝

⎟⎠

The coordinates of the midpoint of Q 3 1,8 8

⎛ −⎜⎝ ⎠

⎞⎟ and R 72,

8⎛ ⎞−⎜ ⎟⎝ ⎠

are 13 3,16 8⎛ ⎞−⎜ ⎟⎝ ⎠

.


Chapter 2 Section 1 Question 4 Page 66 a) Find the coordinates of the midpoint of AB.

( )

( )

1 2 1 2, ,2 2

,2 2

2 6

8 14,2 24

5 9

,7

x x y yx y + +⎛ ⎞= ⎜ ⎟⎝ ⎠

+ +⎛ ⎞= ⎜ ⎟⎝ ⎠⎛ ⎞= ⎜ ⎟⎝ ⎠

=

Find the slope of the median.

2 1

2 1

2 78 4

54

45

y ymx x−

=−

=−

−=

= −

−

The slope of the median is 54

− .


b) Find the coordinates of the midpoint of FG.

( )

( )

1 2 1 2, ,2 2

,2 2

24 10,2 2

1

9 15

5

4 1

,

4

2

x x y yx y

−

+ +⎛ ⎞= ⎜ ⎟⎝ ⎠

+ +⎛ ⎞= ⎜ ⎟⎝ ⎠⎛ ⎞= ⎜ ⎟⎝ ⎠

=


( )( )

2 1

2 1

7121

y ym

5 712 5

x x−

=−

=

=

−− −

−

The slope of the median is 1217

.


( )

( )

1 2 1 2

23.6 79.4 38.0

, ,2 2

,2 2

103 81.8,2 2

51.5

43.

,40

8

.9

x x y yx y + +⎛ ⎞= ⎜ ⎟⎝ ⎠

+ +⎛ ⎞= ⎜ ⎟⎝ ⎠⎛ ⎞= ⎜ ⎟⎝ ⎠

=

The checkpoint should be at (51.5, 40.9).



( )

( )

( )

1 2 1 2, ,2 2

,2 2

8 6,2

7 1 4 10

24,3

x x y yx y

− − −

+ +⎛ ⎞= ⎜ ⎟⎝ ⎠⎛ ⎞+ +

= ⎜ ⎟⎜ ⎟⎝ ⎠−⎛ ⎞= ⎜ ⎟

⎝ ⎠= −

The centre of the circle is at (–4, 3). Chapter 2 Section 1 Question 7 Page 67 Answers may vary. For example: Using The Geometer’s Sketchpad®: Plot the endpoints, and construct the line segment between them. Construct the midpoint of this line segment. Then, select the midpoint and choose Coordinates from the Measure menu. Using Cabri® Jr.: Choose Point from the F2 menu to plot the endpoints. Choose Coord. & Eq. from the F5 menu, and check the placement of the endpoints. Adjust the endpoints if necessary. Choose Segment from the F2 menu, and construct the line segment between the endpoints. Choose Midpoint from the F3 menu, and construct the midpoint. Then, choose Coord. & Eq. again to display the coordinates of the midpoint.


Chapter 2 Section 1 Question 8 Page 67 Find the coordinates of the midpoint, M, of BC.

( )

( )

1 2 1 2, ,2 2

,2 2

6

4 2,2

2 1

0

2

2 2

,

x x y yx y + +⎛ ⎞= ⎜ ⎟⎝ ⎠

+ +⎛ ⎞= ⎜ ⎟⎝ ⎠−⎛ ⎞= ⎜ ⎟

⎠= −

−

⎝


( )

2 1AM

2 1

12

36

y ym

44 2

1x x−

=−−

=

=

=

−−

Substitute m = 12

and the coordinates of one endpoint, say (–2, 1), to find b.

( )11 22

= +−

121

y mx b

b

bb

= +

= − +=

The equation of the median from vertex A is 1 22

y x= + .


Chapter 2 Section 1 Question 9 Page 67 Answers may vary. For example: Using The Geometer’s Sketchpad®: Plot the vertices of ΔABC, and construct the midpoint, M, of side BC. Construct a line through AM. Select the line, and choose Equation from the Measure menu. Using Cabri® Jr.: Choose Point from the F2 menu, and plot the vertices of ΔABC. Choose Coord. & Eq. from the F5 menu, and check the placement of the vertices. Adjust the vertices if necessary. Choose Segment from the F2 menu, and construct the line segment between vertices B and C. Select this line segment and choose Midpoint from the F3 menu. Choose Line from the F2 menu, and construct the line through the midpoint and vertex A. Then, choose Coord. & Eq. again to display the equation of the line.


Chapter 2 Section 1 Question 10 Page 67 a) Find the coordinates of the midpoint, M, of QR.

( )

( )

1 2 1 2, ,2 2

,2

6 32

9 3,2 2

4 5

x x y yx y + +⎛ ⎞= ⎜ ⎟⎝ ⎠⎛ ⎞++

= ⎜ ⎟⎜ ⎟⎝ ⎠⎛

−

⎞= ⎜ ⎟⎝ ⎠


( )

2 1PM

2 1

32

1323

13

y ymx

3 02

9 22

x−

=−

−

=

=

=− −

Substitute m = 313


( )30 213

b= − +

6013

613

y mx b

b

b

= +

= − +

=

The equation of the median from vertex P is 3 613 13

y x= + .


b) Find the coordinates of the midpoint, N, of PR.

( )

( )

1 2 1 2, ,2 2

,2

0 32

3 3,2 2

2 5

x x y yx y + +⎛ ⎞= ⎜ ⎟⎝ ⎠⎛ ⎞++

= ⎜ ⎟

−−⎜ ⎟⎝ ⎠⎛ ⎞= −⎜ ⎟⎝ ⎠


2 1QN

2 1

3 62

3 4215252

3

y ymx x−

=−

−

−=−

=−

=

y mx bb

bb

= +

= +

= +− =

−

6

Substitute m = 3 and the coordinates of one endpoint, say (4, 6), to find b.

( )1

6

3 46 2

The equation of the median from vertex Q is 3 6y x= − .


Chapter 2 Section 1 Question 11 Page 67 Answers may vary. For example: Using The Geometer’s Sketchpad®: a) Plot the vertices of ΔPQR. Construct the midpoint, S, of side QR. Construct a line through points P and S. Select the line, and choose Equation from the Measure menu. b) Construct the midpoint, T, of side PR, and the line though points Q and T. Select the line, and choose Equation from the Measure menu. Using Cabri® Jr.: a) Choose Point from the F2 menu, and plot the vertices of ΔPQR. Choose Coord. & Eq. from the F5 menu, and check the placement of the vertices. Adjust the vertices if necessary. Choose Segment from the F2 menu, and construct the line segment between vertices Q and R. Select this line segment, and choose Midpoint from the F3 menu. Choose Line from the F2 menu, and construct the line through the midpoint and vertex P. Then, choose Coord. & Eq. again to display the equation of the line. b) Use the method in part a) to construct the midpoint T of side PR and the line through points Q and T. Then, choose Coord. & Eq. from the F5 menu to display the equation of the line. Chapter 2 Section 1 Question 12 Page 67

( ) 1 2 1 2, ,2 2

,2 23

2

,

3

22

a a

x x y

b b

yx y

ba

+ +⎛ ⎞= ⎜ ⎟⎝ ⎠

+ +⎛ ⎞= ⎜ ⎟⎝ ⎠⎛ ⎞= ⎜ ⎟⎝ ⎠

These coordinates are the mean of the x-coordinates of the endpoints and the mean of the y-coordinates of the endpoints.


Chapter 2 Section 1 Question 13 Page 67 a)

( )

( )

1 2 1 2

1 1

, ,2 2

6 5,2

x

24 ,2

x y yx y

x y

+ +⎛ ⎞= ⎜ ⎟⎝ ⎠

+ +⎛ ⎞= ⎜ ⎟⎝ ⎠

Set up equations to solve for the coordinates. 1

1

1

6 42

6 82

x

xx

+=

+ ==

1

1

1

5 22

5 41

y

yy

+=

+ == −

The coordinates of the other endpoint are D(2, –1). b) Answers may vary. For example: Refer to part a). Let the coordinates of the other endpoint be D(x1, y1). Solving the equations gives x1 = 2 and y1 = –1. Alternative method: Since the run from C to M is –2, subtract 2 from the x-coordinate of M to find the x-coordinate of D. Since the rise from C to M is –3, subtract 3 from the y-coordinate of M to find the y-coordinate of D. c) Answers may vary. Substitute the coordinates of points C and D into the midpoint formula to confirm that M is the midpoint of CD. Chapter 2 Section 1 Question 14 Page 67 The midpoint of the diameter is at (0, 0). Let the coordinates of the other endpoint be (x1, y1).

( )

( ) ( )

1 2 1 2

1 1

, ,2 2

3 4, ,00−

2 2

x x y yx y

x y

+ +⎛ ⎞= ⎜ ⎟⎝ ⎠⎛ ⎞+ +

= ⎜ ⎟⎝ ⎠


1

1

3 02

3 03

x

xx

−=

− ==

1

1

1

4 02

4 04

y

yy

+=

+ == −

The coordinates of the other endpoint are (3, –4).


Chapter 2 Section 1 Question 15 Page 67 a) Case 1: The centre is D, and the endpoint is E.

( )

( ) ( )

1 2 1 2

1 1

, ,2 2

1 2, ,42−

2 2

x x y yx y

x y

+ +⎛ ⎞= ⎜ ⎟⎝ ⎠⎛ ⎞+ +

= ⎜ ⎟⎝ ⎠


1

1

1 22

1 45

x

xx

−=

− ==

1

1

1

2 42

2 86

y

yy

+=

+ ==

Possible coordinates for the endpoint are (5, 6). Case 2: The centre is E, and the endpoint is D.

( )

( )

1 2 1 2

1 1

, ,2 2

2 4

x

, ,212 2

x y yx y

x y

+ +⎛ ⎞= ⎜ ⎟⎝ ⎠

+ +⎛−

⎞= ⎜ ⎟⎝ ⎠


1

1

2 12

2 24

x

xx

+= −

+ = −= −

1

1

1

4 22

4 40

y

yy

+=

+ ==

Possible coordinates for the endpoint are (–4, 0). b) There are two possible answers for part a) because either D or E could be the centre of the circle.


Chapter 2 Section 1 Question 16 Page 67 Find the coordinates of the midpoint of the line segment PQ.

( )

( )

1 2 1 2

5 3 2, 6

, ,2 2

2 21,2

x x y yx y + +⎛ ⎞= ⎜ ⎟⎝ ⎠⎛ ⎞= ⎜ ⎟⎝ ⎠

= −

+ +− −

Find the slope of the line through P and Q.

( )( )

2 1

2 1

881

y ym

6 23 5

x x−

=−

=

=

=

2 11

y mx bb

bb

= +

= +=

−−− −

2 1 1= +−−

The slope of a line perpendicular to the line segment PQ is –1. Substitute m = 1 and the coordinates of the midpoint to find b.

( )

The equation of the right bisector of PQ is 1y x= − + .


Chapter 2 Section 1 Question 17 Page 67 a) Answers may vary. For example: Any point on the right bisector of a line segment is equidistant from the endpoints. Therefore, points on the right bisector of the line segment joining the two towns are possible locations for the relay tower. b) Find the midpoint of the line segment.

( )

( )

1 2 1 2

2 10 6 0

, ,2 2

,2 2

6,3

x x y yx y + +⎛ ⎞= ⎜ ⎟⎝ ⎠

+ +⎛ ⎞= ⎜ ⎟⎝ ⎠

=

Find the slope of the line segment.

2 1

2 1

6

0 610 2

834

y ymx x−

=−

=−

−=

= −

−

The slope of a line perpendicular to the line segment is 43

.

Substitute m = 43

and the coordinates of the midpoint to find b.

( )433

= +6

3 85

y mx b

b

bb

= +

= +− =

The equation of the right bisector is 4 53

y x= − .


Chapter 2 Section 1 Question 18 Page 67 Answers may vary. For example: Using The Geometer’s Sketchpad®: Plot the points A(2, 6) and B(10, 0). Construct the line segment AB and the midpoint of AB. Then, construct a perpendicular line through the midpoint. Select the perpendicular line, and choose Equation from the Measure menu. Using Cabri® Jr.: Choose Segment from the F2 menu, and plot the endpoints at points (2, 6) and (10, 0). Use Coord. & Eq. from the F5 menu to check the placement of the endpoints, and adjust them if necessary. Select the line segment, and choose Midpoint from the F3 menu. Choose Perp. from the F3 menu, and construct the perpendicular line through the midpoint. Then, choose Coord. & Eq. again to display the equation of the line. Chapter 2 Section 1 Question 19 Page 67 a)


b) Find the coordinates of the midpoint, M, of BC.

( )

( )

( )

1 2 1 2, ,2 2

,2 2

6,3

8 24 8

x x y yx y + +⎛ ⎞= ⎜ ⎟⎝ ⎠⎛ ⎞++

= ⎜ ⎟⎜ ⎟⎝ ⎠

=

−


2 1AM

2 1

3 06 238

y ymx x−

=−

=

=

−+

Substitute m = 38

and the coordinates of one endpoint, say (6, 3), to find b.

( )338

= +6

934

34

y mx b

b

b

b

= +

= +

=

The equation of the median from vertex A is 3 38 4

y x= + .


c) Find the slope of the line segment BC.

( )

2 1

2 1

2

1045

y ym

8 28 4

x x−

=−

=−

=

=

− −

The slope of a line perpendicular to BC is 25

− .

Substitute m = 25

− and the coordinates of the midpoint to find b.

( )1235

275

y mx b

b

b

= +

= − +

=

23 65

b= − +

The equation of the right bisector of BC is 2 25 5

y x= − +7 .

d) Answers may vary. For example: Check that the slopes and y-intercepts on the drawing match those in the equations.


Chapter 2 Section 1 Question 20 Page 68 a), b), e)

c) Since U is the midpoint of PR, RU = UP = 12

PR. Since ST joins the midpoints of two sides of

ΔPQR, ST = 12

PR. Therefore, ST = RU = UP. Similarly, UT = PS = SQ and RT = TQ = US.

Therefore, ΔRUT ΔUPS ΔSTU ≅ ≅ ≅ ΔTSQ (side-side-side).

d) The area of ΔSTU is 14

the area of ΔPQR.

f) The area of one of the smallest triangles is 14

the area of ΔSTU and 116

the area of ΔPQR.


Chapter 2 Section 1 Question 21 Page 68 b) Answers may vary. For example: Join the midpoints of the sides of an equilateral triangle to form four equilateral triangles inside the original triangle. Shade the centre triangle. For each of the other three triangles, repeat the process of joining the midpoints to form smaller similar triangles, and shade the centre triangle. The procedure works with any triangle. The area relationships are the same as shown in question 20 since the line segment joining the midpoints of two sides of any triangle is half the length of the third side. c) d) Answers may vary. For example: Sierpinski’s triangle is a fractal since all of the smaller triangles in each step are similar to the original triangle. Chapter 2 Section 1 Question 22 Page 68

ED is 18

of BC. If ED = 2, then BC is 8 × 2, or 16 units.

Chapter 2 Section 1 Question 23 Page 68 a) The run is 11 – 2 = 9. The rise is 19 – 1 = 18. The required points are (5, 7) and (8, 13). See part b) for the explanation. b) Answers may vary. For example:

For the first dividing point, add 13

of the run to the x-coordinate of the first endpoint and add 13

of the rise to the y-coordinate of the first endpoint. For the second dividing point, add 23

of the

run to the x-coordinate of the first endpoint and add 23

of the rise to the y-coordinate of the first

endpoint.


Chapter 2 Section 1 Question 24 Page 68 a) Use grid paper or dynamic geometry software to plot the given midpoints. A line segment joining the midpoints of two sides of a triangle is parallel to the third side. Use this property to draw the sides of the triangle. b) Midpoint of AB: Midpoint of BC: Midpoint of AC:

( )

( )

1 2 1 2

1 1 2

, ,2 2

,2 2

0,2

6

x x y yx y + +⎛ ⎞= ⎜ ⎟⎝ ⎠

+ +⎛ ⎞⎜

−−= ⎟⎝ ⎠

=

( )

( )

1 2 1 2, ,2 2

,2 2

2

1 3 6 2

,4

x x y yx y + +⎛ ⎞= ⎜ ⎟⎝ ⎠

+ +⎛ ⎞= ⎜ ⎟⎝ ⎠

=

( )

( )

1 2 1 2

1 3 2

, ,2 2

,2 2

1,0

2

x x y yx y + +⎛ ⎞= ⎜ ⎟⎝ ⎠

+ +⎛ ⎞⎜

−−= ⎟⎝ ⎠

= The midpoint formula correctly predicts the coordinates of the midpoints.



a) ( )

( )

1 2 1 2 1 2, , , ,2 2 2

, ,2 2 2

4,

2 6 3 7 1 5

5,3

x x y y z zx y z + + +⎛ ⎞= ⎜ ⎟⎝ ⎠

+ + +⎛ ⎞= ⎜ ⎟⎝ ⎠

=

b) The coordinates of the midpoint of a line in three dimensions are given by the formula:

( ) 1 2 1 2 1 2, , , ,2 2 2

x x y y z zx y z + + +⎛ ⎞= ⎜ ⎟⎝ ⎠

Chapter 2 Section 1 Question 26 Page 68 Answers may vary. For example: All of the points equidistant from the first two towns lie on the right bisector of the line segment joining the two towns. Similarly, all of the points equidistant from the second and third towns lie on the right bisector of the line segment joining them. The point of intersection of these two right bisectors is the only location equidistant from all three towns. Chapter 2 Section 1 Question 27 Page 69 a) Answers may vary. For example: Latitude and longitude are not linear coordinates since the distance between lines of longitude decreases as the distance from the equator increases. The midpoint formula is accurate only for Cartesian coordinates.


Chapter 2 Section 1 Question 28 Page 69 Explanations may vary. a) Sometimes true: Line segments can bisect each other without being equal in length. b) Never true: Parallel lines have no points in common. c) Always true: The midpoint is the only point that is both on the line segment and equidistant from the endpoints. d) Sometimes true: The midpoint of a line segment is equidistant from the endpoints, but so is every other point on the right bisector of the line segment. Chapter 2 Section 1 Question 29 Page 69

( )

( )

1 2 1 2, ,2 2

, ,2

3 43d + +−112

x x yy

c

x y+ +⎛ ⎞= ⎜ ⎟⎝ ⎠⎛=

⎞⎜ ⎟⎝ ⎠

Set up equations to solve for the variables. 3 11

27

d +=

=

4 32

4 610

c

cc

− +=

− + ==

c = 10 and d = 7. Chapter 2 Section 1 Question 30 Page 69 There are 3 vowels in magnetic. There are 3 choices for the last letter. There are 7 choices for the first letter, 6 for the second, 5 for the third, and 4 for the fourth. Number of arrangements 7 6 5 4 3

2520= × × × ×=

Answer D Chapter 2 Section 1 Question 31 Page 69

5 5

6

2 2 32 3642

+ = +=

=

2

The value of x is 6. Answer C


Chapter 2 Section 2 Length of a Line Segment Chapter 2 Section 2 Question 1 Page 77 Estimates will vary.

a) ( ) ( )

( ) ( )

2 22 1 2 1

2 2

2 2

3 2 5 1

1 4

1 16

17

d x x y y= − + −

= − + −

= +

= +

=

The length of line segment AB is 17 units.

b) ( ) ( )

( ) ( )

( )

2 22 1 2 1

2 2

22

8

4 1

16 1

17

4 1 0

d x x y y= − + −

= − + −

= + −

= +

=

−

The length of line segment EF is 17 units.

c) ( ) ( )

( )) ( )( )(

2 22 1 2 1

2 2

2

8 2

64 4

68

d x x y y= − + −

= −

= +

= +

=

2 2

56 3+ − −− −

The length of line segment GH is 68 units.



a) ( ) ( )

( )( ) ( )( )

2 22 1 2 1

22+ −−

2

2 2

4 6 3

10 5

100 25

125

d x x y y= − + −

= − −

=

+

=

+

=


b) ( ) ( )

( )( ) ( )

( )

2 22 1 2 1

2+

2

229 3

81 1

7 3

90

2

6

0

9

d x x y y

− −

= − + −

= − −

= + −

= +

=

The length of line segment CD is 90 units.

c) ( ) ( )

( )( ) ( )( )

2 22 1 2 1

2 24 4

16 16

32

d x x y y= − + −

= −

= +

= +

=

2 2

1 5 2 6− + −−− −

The length of line segment EF is 32 units.

d) ( ) ( )

( ) ( )

( )

2 22 1 2 1

2 2

22

8 0 1 5

8 6

64 36

10010

d x x y y= − + −

= − + −

= + −

= +

==

−

The length of line segment GH is 10 units.



a) ( ) ( )

( ) ( )

( ) ( )

2 22 1 2 1

2 2

2 2

4.5 2.1

6.6 13.0

14.

4.7 3

6

8.

d x x y y

− −

= − + −

= − + −

= − + −

The length of line segment JK is about 14.6 units.

b) ( ) ( )

( )( ) ( )( )

2 22 1 2 1

2 2

2 2

11.6 .

15.8 14.321

2

3

.2

.

9

d x x y y= − + −

= − +

+=

−

4 5.1− −

The length of line segment LM is about 21.3 units.

c) ( ) ( )

( )

2 22 1 2 1

2 2

22

3 1 5 52 2 2

1 5

2−

1 25

26

d x x y y= − + −

⎛ ⎞ ⎛ ⎞= − + −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

= + −

= +

=

The length of line segment NP is 26 units. Chapter 2 Section 2 Question 4 Page 77

( ) ( )

( ) ( )

( )

2 22 1 2 1

2 2

22

4 1 3

3 4

9 16

7

255

d x x y y= − + −

= − + −

= + −

= +

==

The stores are 5 km apart.



a) ( ) ( )

( ) ( )

( )

2 22 1 2 1

2 2

22

8

8 4

64 16

80

0 1 5

d x x y y= − + −

= − + −

= + −

= +

=

( ) ( )

( ) ( )

( )

2 22 1 2 1

2 2

22

8 6 1 11

2 10

2 100

102

d x x y y= − + −

= − + −

= + −

= +

=

The school at (0, 5) is closer to Jordan's house. b) Answers may vary. You could make a scale diagram and measure the distances with a ruler, or use geometry software to plot the points and measure the distances between them.



a) ( ) ( )

( )) ( )( )(

2 22 1 2 1

2 28 6

64 36

10010

2

d x x y y= − + −

= −

= +

= +

==

)

2 2

6 5 1+ −− −

The length of side AB is 10 units.

( ) (

( ) ( )( )( )

2 22 1 2 1

21−2

2 2

2 10 5

8 6

64 36

10010

d x x y y= − + −

= − + −

= − +

= +

==

The length of side AC is 10 units.

( ) ( )

( ) ( )( )( )

2 22 1 2 1

2 216 0

256 0

216

56

d x x y y= − + −

=

= − +

= +

==

226 10 11− + − −− −

The length of side BC is 16 units. b) The perimeter of the triangle is 10 + 10 + 16, or 36 units. c) ΔABC is isosceles.



a) ( ) ( )

( ) ( )

( )

2 22 1 2 1

2 2

2 2

1 1 0 0

2 0

4 0

42

d x x y y= − + −

= − + −

= − +

= +

==

−

The length of side DE is 2 units.

( ) ( )

( ) ( )( ) ( )

2 22 1 2 1

22

221 3

1 3

1 0 3

2

0

4

d x x y y= − + −

= − + −

= − + −

= +

==

−

The length of side DF is 2 units.

( ) ( )

( ) ( )( )

2 22 1 2 1

22

221 3

1 3

42

1 0 0 3

d x x y y= − + −

= − + −

= + −

= +

==

The length of side EF is 2 units. Since all three sides have the same length, ΔDEF is equilateral. b) Answers may vary. For example: Any enlargement of ΔDEF, such as (–2, 0), (2, 0), and (0, 2 3 ), or any translation, such as (0, 0), (2, 0) and (1, 3 ), will result in another equilateral triangle.


Chapter 2 Section 2 Question 8 Page 78 Find the coordinates of the midpoint, M, of side KL.

( ) 1 2 1 2

3 1

, ,2 2

2 3,2 251,2

x x y yx y + +⎛ ⎞= ⎜ ⎟⎝ ⎠

+ +−⎛ ⎞= ⎜ ⎟⎝ ⎠⎛ ⎞= −⎜ ⎟⎝ ⎠

Find the length of the median.

( ) ( )

( ))(

( )

2 22 1 2 1

2

22 91

2

8114

854

d x x y y= − + −

⎛ ⎞= ⎜ ⎟⎝ ⎠

⎛ ⎞= − + −⎜ ⎟⎝ ⎠

= +

=

2 52 1 22

− + −− − −

The length of the median from vertex J is 854

units.

Chapter 2 Section 2 Question 9 Page 78 Answers may vary. For example: Using The Geometer’s Sketchpad®: Plot the points J, K, and L. Construct line segment KL and its midpoint, M. Then, construct and measure the length of line segment JM. Using Cabri® Jr.: Choose Triangle from the F2 menu, and construct ΔJKL. Choose Coord. & Eq. from the F5 menu, and display the coordinates of the vertices. Adjust the vertices if necessary. Choose Midpoint from the F3 menu, and select side KL. Choose Segment from the F2 menu, and construct the line segment from the midpoint to vertex J. Choose Measure/D. & Length from the F5 menu, and select the median.



( ) ( )

( )( ) ( )( )

2 22 1 2 1

2

2 2

4 2 4

6 6

36 36

72

d x x y y= − + −

= −

= +

= +

−

=

2

2+ −−

The length of side RS is 72 units.

( ) ( )

( ) ( )( )( )

2 22 1 2 1

22

2 2

4 10 4

6 6

36 36

72

2−

d x x y y= − + −

= − +

= − +

= +

=

−

The length of side RT is 72 units.

( ) ( )

( ) ( )( )( )

2 22 1 2 1

22

2 212 0

144 0

2

2

10 2

1

2

d x x y y

− −

= − + −

= − + −

= − +

= +

−

= The length of side ST is 12 units. ST is the hypotenuse. Use the other two sides to calculate the area.

72

12121 72236

72

A bh=

= ×

= ×

=

The area of ΔRST is 36 square units.


Chapter 2 Section 2 Question 11 Page 78 Answers may vary. For example: Using The Geometer’s Sketchpad®: Construct the triangle with vertices R, S, and T. Then, select and measure the interior of ΔRST. Using Cabri® Jr.: Choose Triangle from the F2 menu, and construct ΔRST. Choose Coord. & Eq. from the F5 menu, and display the coordinates of the points. Adjust the position of a vertex if its coordinates are not correct. Choose Measure/Area from the F5 menu, and select ΔRST.



( ) ( )

( )( ) ( )

( ) ( )

2 22 1 2 1

2

2 23 6

9 36

45

5

d x x y y= − + −

= −

= − + −

= +

=

2

5 2 1− + −− −

The length of line segment CA is 45 units.

( ) ( )

( )) ( )( ( )

2 22 1 2 1

2 23 6

9 36

45

d x x y y= − + −

=

= +

= +

=

2 2

5 8 1 7− + −− − − −

The length of line segment CB is 45 units.

( ) ( )

( )( ) ( )( )

2 22 1 2 1

2 2

2 2

2 8 5

6 12

36 144

180

2 4

7

5

d x x y y

− − −

= − + −

= − + −

= +

= +

=

=

The length of line segment AB is 2 45 units.

Since CA = CB = 12

AB, C is the midpoint of line segment AB.



a) ( )

( )

1 2 1 2, ,2 2

,2 2

51

72 4

,2

2

x x y yx y + +⎛ ⎞= ⎜ ⎟⎝ ⎠⎛ ⎞++

= ⎜ ⎟⎝ ⎠⎛

−−

⎞= ⎜ ⎟⎝ ⎠

The coordinates of the midpoint, M, of KL are 51,2

⎛ ⎞⎜ ⎟⎝ ⎠

.

b)

( ) ( )

( )

( )

2 22 1 2 1

22

22 93

2

8194

11

52 1 72

74

d x x y y= − + −

⎛ ⎞= − + −⎜ ⎟⎝ ⎠

⎛ ⎞= − + ⎜ ⎟⎝ ⎠

=

=

−

+

The length of line segment KM is 1174

units.

( ) ( )

( )

2 22 1 2 1

22

22

54 1

932

8194

11

2

74

2

d x x y y= − + −

⎛ ⎞= − + −⎜ ⎟⎝ ⎠

⎛ ⎞= + −⎜ ⎟⎝ ⎠

=

−

+

=

The length of line segment LM is 1174

units.


( ) ( )

( ) ( )( )( )

2 22 1 2 1

2 26 9

36 81

117

d x x y y= − + −

=

= − +

= +

=

222 4 7 2− + −− −

The length of line segment KL is 117 units.

Since KM = LM = 12

KL, M is the midpoint of KL.


( ) ( )

( ) ( )

( ) ( )

2 22 1 2 1

2 2

2 2

2 10 2 17

8 15

64 225

28917

d x x y y= − + −

= − + −

= − + −

= +

==

The pipe is 17 m long. The cost is 17 × $3.15, or $53.55.


Chapter 2 Section 2 Question 15 Page 78 a) b) Midpoint of PQ: Midpoint of PR:

( ) 1 2 1 2

3 5

, ,2 2

,2 231,2

4 1

x x y yx y + +⎛ ⎞= ⎜ ⎟⎝ ⎠

+ +⎛ ⎞= ⎜ ⎟⎝ ⎠⎛ ⎞= −

( ) 1 2 1 2

− −

⎜⎝

⎟⎠

3 2 , 4 7

, ,2 2

2 21 3,2 2

x x y yx y

−

+ +⎛ ⎞= ⎜ ⎟⎝ ⎠⎛ ⎞= ⎜ ⎟⎝ ⎠⎛ ⎞= −⎜ ⎟⎝ ⎠

−+ +

The coordinates of the midpoint of PQ are S 31,2

⎛ −⎜⎝ ⎠

⎞⎟ . The coordinates of the midpoint of PR are

T 1 3,2 2

⎛ ⎞−⎜ ⎟⎝ ⎠

.


c)

( ) ( )2 22 1 2 1

22

223 3

2

9 94454

1

1 3 312 2 2

⎛ ⎞⎞ ⎛ ⎞− + −⎟ ⎜ ⎟⎜ ⎟− −

452

d x x y y= − + −

⎛= ⎜⎝ ⎠ ⎝ ⎠⎝ ⎠

⎛ ⎞= − +⎜ ⎟⎝ ⎠

= +

=

=

The length of ST is 1 452

units.

( ) ( )

( ) ( )

( )

2 22 1 2 1

2 2

2 2

2 5 7

3 6

9 36

45

1

d x x y y= − + −

= − + −

= − +

= +

=

The length of QR is 45 units.

Therefore, ST = 12

QR.


d)

2 1ST

2 1

3 32 2

121

3322

y ymx x−

=−

=⎛ ⎞⎜ ⎟⎝

−=

−

− −

= −

−

⎠

2 1QR

2 1

6

2

27 1

5

3

y ymx x−

=−

=−

=−

= −

−

Since the slopes are the same, ST is parallel to QR.


e) Answers may vary. For example: Using The Geometer's Sketchpad®, construct ΔPQR, and the midpoint triangle, ΔSTU. Measure the lengths of the sides. The sides of the midpoint triangle are each half the length of the corresponding side on the main triangle. Hence, the triangles are similar.


Chapter 2 Section 2 Question 16 Page 78 Answers may vary. For example: Using The Geometer’s Sketchpad®: a) Construct the triangle with vertices P, Q, and R. b) Construct the midpoint of PQ and of PR. Then, display the coordinates of the midpoints. c) Measure and compare the lengths of ST and QR. d) Measure and compare the slopes of ST and QR. e) Measure and compare either the side lengths or the angles of ΔPQR and ΔSTU, where U is the midpoint of QR. Using Cabri® Jr.: a) Choose Triangle from the F2 menu, and construct ΔPQR. Choose Coord. & Eq. from the F5 menu, and display the coordinates of the vertices. Adjust the vertices if necessary. b) Choose Midpoint from the F3 menu, and construct the midpoint of PQ and of PR. Choose Coord. & Eq. from the F5 menu, and select the midpoints. c) Choose Measure/D. & Length from the F5 menu. Then, select ST and QR. d) Choose Measure/Slope from the F5 menu. Then, select ST and QR. e) Use the Measure options in the F5 menu to compare either the side lengths or the angles of ΔPQR and ΔSTU, where U is the midpoint of QR.



a) ( ) ( )

( ) ( )

2 22 1 2 1

2 23978 1142 2520

2851

2232

d x x y y= − + −

= − + −

The distance from Edmonton to Ottawa is about 2851 km.

( ) ( )

( ) ( )

2 22 1 2 1

2 21268 1015

504

2540 2104

d x x y y= − + −

= − + −

The distance from Montréal to Toronto is about 504 km.

( ) ( )

( ) ( )

2 22 1 2 1

2 23978 1268 2520

2710

2540

d x x y y= − + −

= − + −

The distance from Edmonton to Toronto is about 2710 km. b) Answers may vary. For example: Flying distances are about 2840 km from Edmonton to Ottawa, 505 km from Montréal to Toronto, and 2705 km from Edmonton to Toronto. The telephone coordinate system gives distances that are close to the flying distances.


Chapter 2 Section 2 Question 18 Page 79 a) Two of the coordinates on the triangle are (1, 0) and (2, 0). Use the Pythagorean theorem to determine the height of the

triangle as 32

. The coordinates of the this point are 3 3,2 2

⎛ ⎞−⎜ ⎟

⎝ ⎠.

b) Koch snowflakes are fractals. Explanations may vary. The sides inserted in each step are similar to two sides in the original triangle and the angle at each new point of the snowflake is equal to the angles in the original triangle. Chapter 2 Section 2 Question 19 Page 79 Solutions for Achievement Checks are shown in the Teacher Resource. Chapter 2 Section 2 Question 20 Page 79 a) A possible value for x is 2. b) This is the only possible answer. Use the distance formula.

( ) ( )

( ) ( )

( )( )( )

2 22 1 2 1

2 2

2

2

2

5 = −2 6 1

5 2 25

25 2 25

0 20 2

2

d x x y y

x

x

xx

x

x

= − + −

+ −

= − +

= − +

= −

= −=


Chapter 2 Section 2 Question 21 Page 79 Answers may vary. Use the origin as one endpoint. i) Use (1, 1) as the other endpoint.

( ) ( )

( ) ( )

2 22 1 2 1

2 21 0

1 1

2

1 0

d x x y y= − + −

= − + −

= +

=

ii) Use (2, 1) as the other endpoint.

( ) ( )

( ) ( )

2 22 1 2 1

2 22 0

4 1

5

1 0

d x x y y= − + −

= − + −

= +

=

iii) Use (3, 2) as the other endpoint.

( ) ( )

( ) ( )

2 22 1 2 1

2 23 0

9 4

2

1

0

3

d x x y y= − + −

= − + −

= +

=

iv) Use (5, 4) as the other endpoint.

( ) ( )

( ) ( )

2 22 1 2 1

2 2

25 1

0

4

4

6

1

5 0

d x x y y= − + −

= − + −

= +

=


Chapter 2 Section 2 Question 22 Page 79 Answers may vary. Any four of the points are acceptable. a) (5, 0), (4, 3), (3, 4), (0, 5), (–3, 4), (–4, 3), (–5, 0), (–4, –3), (–3, –4), (0, –5), (3, –4), (4, –3) b) (7, 1), (6, 4), (5, 5), (2, 6), (–1, 5), (–2, 4), (–3, 1), (–2, –1), (–1, –3), (2, –4), (5, –3), (6, –2) c) (–15, –2), (–13, 4), (–11, 6), (–5, 8), (1, 6), (3, 4), (5, –2), (3, –8), (1, –10), (–5, –12), (–11, –10), (–13, –8) Chapter 2 Section 2 Question 23 Page 79 Use a coordinate grid to map the path. The present is at (–2, 11), a distance of 11.2 m from the origin.


Chapter 2 Section 2 Question 24 Page 79 The letters of net must appear in order. If they form the first three letters, there are 5×4 = 20 ways to fill the remaining two letters. The word may also appear in the middle three spaces, or the last three spaces, for a total of 20 + 20 + 20 = 60 arrangements. Answer A. Chapter 2 Section 2 Question 25 Page 79 Answers may vary. For example: Substitute the Pythagorean theorem into the area formula for the large semicircle.

( )2 2

2 2

1 12 2

1 12 2

a b c

b c

π π 2

π π

= +

= +


Chapter 2 Section 3 Apply Slope, Midpoint, and Length Formulas Chapter 2 Section 3 Question 1 Page 89

The slope of the line shown is –2. The slope of line segment AB is 12

.

Substitute m = 12

and the coordinates of the known endpoint, (1, 0), to

find b.

( )102

= +1

12

y mx b

b

b

= +

− =

The equation of a line containing the line segment AB is 1 12 2

y x= − .

Chapter 2 Section 3 Question 2 Page 89 Answers may vary. For example: If the triangle contains a right angle, the slopes of two of the sides are negative reciprocals of each other and the lengths of the sides are related by the Pythagorean theorem.



b)

( )

2 1CD

2 1

23

y ym

4 21 2

x x

=

−=

−

=

−− −

2 1CE

2 1

1134

2

3

32

y ym − =x x−−−

=

−=

= −

The slopes are negative reciprocals. Therefore, ∠C is a right angle. Chapter 2 Section 3 Question 4 Page 89 Find the coordinates of the midpoint, M, of side JL.

( )

( )

1 2 1 2, ,2 2

,2 2

2

1 3 2 4

,3

x x y yx y + +⎛ ⎞= ⎜ ⎟⎝ ⎠

+ +⎛ ⎞= ⎜ ⎟⎝ ⎠

=


( ) ( )

( ) ( )

2 22 1 2 1

2 24 2

4 9

0

1

3

3

d x x y y= − + −

= − + −

= +

=

The length of the median from vertex K is 13 .


Chapter 2 Section 3 Question 5 Page 89 a) Find the coordinates of M and N.

( )

( )

( )

1 2 1 2M , ,

6 63

2 2

,2

1 0

12

,

x x y yx y + +⎛ ⎞= ⎜ ⎟⎝ ⎠⎛ ⎞++

= ⎜ ⎟⎝ ⎠

= −

( )

− −

( )

1 2 1 2N , ,2 2

,2 2

1,4

6 23 5

x x y yx y + +⎛ ⎞= ⎜ ⎟⎝ ⎠

+ +⎛ ⎞= ⎜ ⎟⎝ ⎠

=

−

Find the slopes of MN and QR.

( )

2 1MN

2 1

2

y ym

4 01 1

x x−

=

−=

=

−− −

( )

2 1QR

2 1

842

y ym

2 65 1

x x−

=−

−

=

− −

=

=Since the slopes are the same, MN is parallel to QR. b) Find the lengths of MN and QR.

( ) ( )

(( ) ( ))

2 22 1 2 1

2

MN

4 16

2

2

0

5

1

0

1 42

x x y y

−

= − + −

= − −

= +

=

=

( ) ( )

+

( ) ( )( )

2 22 1 2 1

25 1

QR

16 64

80

4 5

2 6

x2

x y y

−

= − + −

= − + −

= +

=

=

)

The length of MN is half the length of QR. Chapter 2 Section 3 Question 6 Page 89 Find the lengths of TU and TV.

( ) (

( ) ( )( )

2 22 1 2 1

223 2 5

T

1 3

7

1

U

6

3

x x y y= − + −

= − + −

= +

=

−

( ) ( )

( ) ( )( )

2 22 1 2 1

22

TV

25

2

05

x x y y= − + −

= − +

=

−

+=

3 11 −−−

The lengths are not the same. T does not lie on the right bisector of line segment UV.


Chapter 2 Section 3 Question 7 Page 89 a) Find the slopes of the four sides of the quadrilateral.

2 1OP

2 1

5 03 053

y ymx x−

=−

=

=

−−

2 1RQ

2 1

6 18 553

y ymx x−

=−

=

=

−−

2 1PQ

2 1

6 58 315

y ymx x− 2 1

OR2 1

=−−−

=

=

1 05 015

y ymx x−

=−

=

=

−−

OP is parallel to RQ, and PQ is parallel to OR. OPQR is a parallelogram. b) Answers may vary. For example: Use geometry software to construct OPQR. Measure the slope of each side. These slopes show that the opposite sides are parallel. Chapter 2 Section 3 Question 8 Page 89

a) ( )

( )

1 2 1 2

3 9 5

, ,2 2

,2 2

3,6

7

x x y yx y + +⎛ ⎞= ⎜ ⎟⎝ ⎠

+ +⎛ ⎞= ⎜ ⎟⎝ ⎠

=

−

The coordinates of the centre of the circle are (3, 6).

b) ( ) ( )

( )( ) ( )

2 22 1 2 1

23 6+−

23

36 1

37

5

d x x y y= − + −

= − −

= +

=

The radius of the circle is 37 units. Chapter 2 Section 3 Question 9 Page 89 Find the lengths of the three sides of ∆ABC.

( ) ( )

( )( ) ( )

2 22 1 2 1

2

AB

4 36

40

4

x x y y= − + −

= −

= +

=

21 3 2− + −− −

( ) ( )

( )( ) ( )( )

2 22 1 2 1

3

16 16

32

2

x x y y= − + −

= −

= +

−

=

( ) ( )2 2

1 2+ −−

BC

( )( ) ( )

2 22 1 2 1

23

AC

36 4

40

4

x2

3 2

x y y= − + −

= − −

= +

=

+−

Since AB = AC, ΔABC is isosceles.



The given line has a slope of 2. A line perpendicular to this line has a slope of 12

− .

( )122

= +− 5

522

92

y mx b

b

b

b

= +

= − +

=

The equation of the line perpendicular to the given line through the given point is 1 92 2

y x= − + .

Find the point of intersection. y = 2x + 1

y = 12

− x + 92

Substitute 2x + 1 for y in equation . 1 92 21 92 2

1 92 12 25 7

7

12x +

2 2

5

y x

x

x x

x

x

= − +

= − +

+ = −

=

=

Substitute x = 75

into equation .

2 1

2 1

14 55 5

75

195

y x= +

⎛ ⎞= +⎜ ⎟⎝ ⎠

= +

=

The point of intersection is 7 19,5 5

⎛ ⎞⎜ ⎟⎝ ⎠

.


Find the distance from the given point to the point of intersection.

( ) ( )2 22 1 2 1

2 2

2 218 95 5

324 8125 2540525

815

95

7 195 25

4

5

.02

d x x y y= − + −

⎛ ⎞ ⎛ ⎞= − + −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

⎛ ⎞ ⎛ ⎞= + −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

= +

=

=

=

The shortest distance from the point (5, 2) to the line 2 1y x= + is 95

.



The given line has a slope of 12

. A line perpendicular to this line has a slope of –2.

( )00 = +2 0y mx b

bb−

= +

=

The equation of the line perpendicular to the given line through the given point is y = –2x. Find the point of intersection.

y = 12

x – 2

y = –2x Substitute –2x for y in equation .

1 221 22

12 225 225 4

45

2x−

y x

x

x x

x

x

x

= −

= −

− − = −

− = −

− = −

=

Substitute x = 45

into equation .

2

2

85

45

y x= −

⎛ ⎞= − ⎜ ⎟⎝ ⎠

= −


⎛ ⎞−⎜ ⎟⎝ ⎠

.


Find the distance from the origin to the point of intersection.

( ) ( )2 22 1 2 1

22

16 6425 25

4 80 05 5

8025165

45

d x x y y= − + −

⎛ ⎞⎛ ⎞ ⎛ ⎞= − + −⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠

= +

=

=

=

−

The shortest distance from the origin to the line 1 22

y x= − is 45

.


Chapter 2 Section 3 Question 12 Page 89 3 5 4 0

5 3 43 45 5

x yy x

y x

+ − == − +

= − +


− . A line perpendicular to this line has a slope of 53

.

( )254

543

= +5

3133

y mx b

b

b

b

= +

= +

− =

The equation of the line perpendicular to the given line through the given point is 5 13 3

y x= −3 .

Find the point of intersection.

y = 35

− x + 45

y = 53

x – 133

Substitute 35

− x + 45

for y in equation .

5 133 35 13 Multiply by 15.3 3

9 12 25 6534 77

77

45

35

x +−

34

y x

x

x xx

x

= −

= −

− + = −− = −

=


Substitute x = 7734

into equation .

3 45 53 45 5231 136170 170

7

95170

1934

734

y x= − +

⎛ ⎞= − +⎜ ⎟⎝ ⎠

= − +

−=

= −


⎛ ⎞−⎜ ⎟⎝ ⎠

.


( ) ( )2 22 1 2 1

22

2 293 15534 34

8649 24 0251156 115632 6741156

32 67434

5

77 195 434 34

.32

d x x y y= − + −

⎛ ⎞⎛ ⎞ ⎛ ⎞= − + −⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠

⎛ ⎞ ⎛ ⎞= +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

= +

=

−

=

The shortest distance from the point (5, 4) to the line 3 5 4 0x y+ − = is32 674

34.


Chapter 2 Section 3 Question 13 Page 89 Find the equation of the line through the given points.

( )

2 1FG

2 1

14

28

y ym

43 5

2x x−

=−−

=

=

=

−−

Substitute m = 14


( )524

134

y mx b

b

b

= +

= − +

=

12 54

b= − +

The equation of the line through the given points is 1 14 4

y x 3= + .



( )4 4 1−−

4 40

y mx bb

bb

= +

= +

− = − +=

The equation of the line perpendicular to 1 14 4

y x 3= + through the given point is y = –4x.



y = 14

x + 134

y = –4x Substitute –4x for y in equation .

1 134 41 13 Multiply by 4.4 4

16 1317 13

13

4x−

17

y

xx

x

x

x

x

= +

= +

− = +− =

−=

Substitute x = 1317

− into equation .

1317

4

4

5217

y x=

⎛ ⎞= − ⎜ ⎟⎝ ⎠

=

−


⎛ ⎞−⎜ ⎟⎝ ⎠

.


( ) ( )2 22 1 2 1

2 2

2 2

13 52417 17

⎞⎞ ⎛ ⎞+ −⎟ ⎜ ⎟⎟− −1

30 12017 17

900 14 400289 28915 300

28990017

3017

7.28

d x x y y= − + −

⎛ ⎛= − ⎜⎜⎝ ⎠ ⎝ ⎠⎝ ⎠

−⎛ ⎞ ⎛ ⎞= +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

= +

=

=

=

The shortest distance from the point E(1, –4) to the line through F(–5, 2) and G(3, 4) is 3017

.


Chapter 2 Section 3 Question 14 Page 89 Find the equation of the line through the given points.

( )

2 1JK

2 1

4 462

− −− −

( )2

4 6−

= − − +

842

y ymx x−

=−

=−

−=

= −

+

= +− =

Substitute m = –2 and the coordinates of one point, say (–6, 4), to find b.

24 128

y x bb

bb

=

The equation of the line through the given points is y = –2x – 8.

The given line has a slope of –2. A line perpendicular to this line has a slope of 12

.

( )122522

1

5

2

y mx b

b

b

b

= +

= +

= +

− =

The equation of the line perpendicular to y = –2x – 8 through the given point is 1 12 2

y x= − .



y = 12

x – 12

y = –2x – 8 Substitute –2x – 8 for y in equation .

1 12 21 1 Multiply by 2.2 2

4 16 15 15

82

3

y x

x

x xx

x

x −−

3−

= −

= −

− − = −− =

= −

2

y x= − −

= − −

= −

Substitute x = –3 into equation .

( )2 82 8

The point of intersection is ( )3, 2− − . Find the distance from the given point to the point of intersection.

( ) ( )

( )) ( )( )(

2 22 1 2 1

2 2

5

8 4

64 16

80

4 5

d x x y y= − + −

= −

= +

= +

=

=

2 2

3 2 2+ −− −

The shortest distance from the point H(5, 2) to the line through J(–6, 4) and K(–2, –4) is 4 5 .


Chapter 2 Section 3 Question 15 Page 90 Answers may vary. For example: Using The Geometer’s Sketchpad®: a) Construct line segment AB and point R. Construct a perpendicular from point R to AB. Construct point D, the point of intersection of the perpendicular and AB. Display the coordinates of point D. Line segment RD represents the shortest route. Measure the length of RD, and multiply by 0.5 to find the length of the side road in kilometres.


b) Construct ΔABC. Measure the angles or compare the slopes of the sides to determine that ACB is a right angle. ∠

c) Construct the midpoint, D, of side AB. Construct line segment CD. Measure and compare the lengths of AB and CD. Rolly: Do you still the GSP


Using Cabri® Jr.: a) Choose Segment from the F2 menu, and construct line segment AB. Choose Coord. & Eq. from the F5 menu, and display the coordinates of the points. Adjust their positions if necessary. Choose Point from the F2 menu, and construct point R. Choose Perp. from the F3 menu, and construct the perpendicular from point R to AB. Choose Coord. & Eq. from the F5 menu, and select point D, the point of intersection of the perpendicular and AB. Line segment RD represents the shortest route. Choose Measure/D. & Length from the F5 menu, and select RD. Multiply the displayed length by 0.5 to find the length of the side road in kilometres. b) Choose Triangle from the F2 menu, and construct ΔABC. Choose Coord. & Eq. from the F5 menu, and display the coordinates of the vertices. Adjust the vertices if necessary. Choose Measure from the F5 menu. Then, choose Angle and measure the angles of ΔABC, or choose Slope and measure the slopes of the three sides. Both sets of measurements show that ∠ACB is a right angle. c) Choose Midpoint from the F3 menu, and select side AB. Choose Segment from the F2 menu, and construct line segment CD. Choose Measure/D. & Length from the F5 menu, and select AB and CD.


Chapter 2 Section 3 Question 16 Page 90 Find the equations of a line through C parallel to AB, and a line through A parallel to BC.

( )

2 1AB

2 1

771

y ym

4 352

x x−

=−

=−

=−

= −

7 16

y mx bb

bb

= +

= +=

−−−

7 1 1= +−−

Substitute m = –1 and the coordinates of point C to find b.

( )

The equation of the line through C parallel to AB is y = –x + 6.

( )

2 1BC

2 1

313

y ymx x−

=−

=

=

=

18

3 5y mx b

bb

b

= +

= +

− +−

==

7 41 2−−−−

3−

Substitute m = 3 and the coordinates of point A to find b.

( )3 15

The equation of the line through A parallel to BC is y = 3x + –18.


Find the intersection of these lines. y = –x + 6 y = 3x + –18 Substitute –x + 6 for y in equation .

6x +−

6−

3 13 1

4 2446

yx

x

88

x

x

= −= −

− = −=

Substitute x = 6 into equation . 66

0

y x= − += +=

The coordinates of D are (6, 0). Checks will vary. A check using geometry software is shown.


Chapter 2 Section 3 Question 17 Page 90 a) Find the coordinates of the midpoint, M, of FG.

( )

( )

1 2 1 2

4 0 4, 4

, ,2 2

2 22,0

x x y yx y + +⎛ ⎞= ⎜ ⎟⎝ ⎠⎛ ⎞= ⎜ ⎟⎝ ⎠

= −

+ +− −


( )

2 1EM

2 1

2412

y ym

2 02 2

x x−

=−

=

−=

= −

− −− −

Substitute m = 12

− and the coordinates of one of the endpoints, say (2, –2), to find b.

( )2 11

122

= +−− 2

y mx b

b

bb

= +

− = − +− =

The equation for the median from vertex E is 1 12

y x= − − .

b) ( ) ( )

( ) ( )( )( )

2 22 1 2 1

22

2 2

2 2 0

4 2

16 4

2

2

0

2 5

d x x y y= − + −

= − + −

= − +

=

=

−−

=

+

The length of the median from vertex E is 2 5 .


Chapter 2 Section 3 Question 18 Page 90 a) b) Find the slope of side EF.

2 1EF

2 1

4 30 474

74

y ymx x−

=−

=

−=−

=

− −−

The required slope of the altitude is 47

− .

Substitute m = 47

− and the coordinates of point D to find b.

( )46 17

= +−−

467

387

y mx b

b

b

b

= +

= +

=

The equation for the altitude from vertex D is 4 37 7

y x= − +8 .


Chapter 2 Section 3 Question 19 Page 90 Answers may vary. For example: Using The Geometer’s Sketchpad®: a) Construct the triangle with vertices D, E, and F. Then, construct the perpendicular from D to EF. b) Select the perpendicular and choose Equation from the Measure menu. Using Cabri® Jr.: a) Choose Triangle from the F2 menu, and construct ∆DEF. Choose Coord. & Eq. from the F5 menu, and display the coordinates of the vertices. Adjust the vertices if necessary. Choose Perp. from the F3 menu, and construct the perpendicular from D to EF. b) Choose Coord. & Eq. from the F5 menu, and select the perpendicular.


Chapter 2 Section 3 Question 20 Page 90 a) Find the slopes of the four sides of the quadrilateral.

( )

2 1PQ

2 1

34

y ymx x−

=−−

=

=8

2 54−−−

2RS

2 1

1

43

43

y ym

2 23 6

x x−

=−

=

−=−

=

− −−

( )

2 1PS

2 1

6834

y ym

2 43 5

x x−

=−

− −

( )

2 1QR

2 1

=

−=

= −

− −2 8

6 26

834

y ymx x−

=−

=−

−=

= −

)

−−

The slopes show that PQ and RS, and PS and QR, are parallel. The slopes show that each pair of adjacent sides is perpendicular. The quadrilateral is a rectangle.

b) ( ) (

( )( ) ( )

( )

2 22 1 2 1

2 2

22

PR

11 2

121 4

125

5 5

6 5 2 4

x x y y= − + −

= − −

= + −

= +

=

=

)

+−

( ) (

( )) ( )(

2 22 1 2 1

2 2

3

QS

5 10

25 100

125

5 5

x x y y= − + −

= −

= +

= +

=

=

2 22 2 8+ −− −

The length of each diagonal is 5 5 . c) Midpoint of PR: Midpoint of QS:

( ) 1 2 1 2

5

, ,2 2

,2 2

1 ,3

6 4

2

2

x x y yx y + +⎛ ⎞= ⎜ ⎟⎝ ⎠

+ +−⎛ ⎞= ⎜ ⎟⎝ ⎠⎛ ⎞= ⎜ ⎟⎝ ⎠

( )

( )

1 2 1 2, ,2 2

,2 2

1

22 3

,32

8

x x y yx y + +⎛ ⎞= ⎜ ⎟⎝ ⎠⎛ ⎞++

= ⎜ ⎟⎝ ⎠⎛

−−

⎞= ⎜ ⎟⎝ ⎠

The coordinates of the midpoint of each diagonal are 1 ,32

⎛ ⎞⎜ ⎟⎝ ⎠

.

d) The diagonals of PQRS bisect each other.


Chapter 2 Section 3 Question 21 Page 90 a) Find the slope of side JK.

( )

2 1JK

2 1

3612

y ym

3 04 2

x x−

=−

=

−=

= −

y mx bb

bb

= +

= +

= +− =

− −− −

8

The required slope of the altitude is 2. Substitute m = 2 and the coordinates of point L to find b.

( )1

8

2 88 6

The equation for the altitude from vertex L is y = 2x – 8. b) Find the equation of the line through JK.

( )10 22

= +−−

0 11

y mx b

b

bb

= +

= +− =

The equation of the line through JK is y = 12

− x – 1.


Find the intersection of the altitude with JK.

y = 12

− x – 1

y = 2x – 8 Substitute 2x – 8 for y in equation .

1 121 1212 8 1 Multiply by 2.2

4 16 25 1

2 8x = −−

4145

y x

x

x x

x xx

x

= − −

−

− = − −

− = − −=

=

Substitute x = 145

into equation .

2 8

2 8

28 405 5125

145

y x= −

⎛ ⎞= −⎜ ⎟⎝ ⎠

= −

= −

The intersection is at D 14 12,5 5

⎛ ⎞−⎜ ⎟⎝ ⎠

.


Find the length of the altitude.

( ) ( )2 22 1 2 1

22

2 2

DL

26 525 5

676 270425 25

338025

14 128 85 5

6765

265

x x y y= − + −

⎛ ⎞⎛ ⎞ ⎛ ⎞= − + −⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠

⎛ ⎞ ⎛ ⎞= +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

= +

=

=

=

−

The length of the altitude from vertex L is 265

.

c) Find the length of side JK.

( ) ( )

( )( ) ( )

( )

2 22 1 2 1

2+ 2

22

JK

6 3

36 9

4

0

5

2

3 5

4 3

x x y y

− −

= − + −

= − −

= + −

= +

=

=

Use the lengths of side JK and the altitude from vertex L to find the area.

( )26 3 55

121239

A bh=

⎛ ⎞= ⎜ ⎟⎝ ⎠

=

The area of ΔJKL is 39 square units.


Chapter 2 Section 3 Question 22 Page 90 Answers may vary. For example: Using The Geometer’s Sketchpad®: a) Construct the triangle with vertices J, K, and L. Construct the perpendicular from L to JK. Construct point M, the point of intersection of the perpendicular and JK. Construct line segment LM. Measure the length of LM. b) Select the perpendicular, and choose Equation from the Measure menu. c) Select the interior of ∆JKL and choose Area from the Measure menu.


Using Cabri® Jr.: a) Choose Triangle from the F2 menu, and construct ∆JKL. Choose Coord. & Eq. from the F5 menu, and display the coordinates of the vertices. Adjust the vertices if necessary. Choose Perp. from the F3 menu, and construct the perpendicular from L to JK. Choose Measure/D. & Length from the F5 menu, and select the endpoints of the altitude. b) Choose Coord. & Eq. from the F5 menu, and select the perpendicular. c) Choose Measure/Area from the F5 menu, and select ΔJKL. Chapter 2 Section 3 Question 23 Page 90

a) The given line has a slope of 12


Substitute m = –2 and the coordinates of point H to find b.

( )17 731

2y mx b

bb−

= +

= +

=

The equation of the line perpendicular to the given line is y = –2x + 31. Find the point of intersection.

y = 12

x + 4

y = –2x + 31 Substitute –2x + 31 for y in equation .

1 421 4 Multiply by 2.2

4 62

2 31x

85 54

10.8

y x

x

x xx

x

= +

− + = +− = −

=

19.4

y x= − +

= − +

=

= +− +

.0 8

Substitute x = 10.8 into equation .

( )12 312 3

The point of intersection is (10.8, 9.4).


b) Find the distance from the house to the point of intersection.

( ) ( )

( ) ( )

2 22 1 2 1

2+

2

8

7 10.8 17 9.

5

4

.

d x x y y= − + −

= − −

Using the given scale, the shortest distance from the house to the trunk cable is 10 × 8.5, or 85 m. Chapter 2 Section 3 Question 24 Page 90 a) Find the equation of the line through the given points.

2 1

2 1

5 76 4

221

y ymx x−

=−

=−

−=

= −

7 411

y mx bb

bb

= +

= +

= − +=

21 6

y mx bb

b

= +

= +

=

−

7 1 4−

8

Substitute m = –1 and the coordinates of one point, say (4, 7), to find b.

( )

The equation of the line through the given points is y = –x + 11. The given line has a slope of –1. A line perpendicular to this line has a slope of 1. Substitute m = 1 and the coordinates of the hikers, to find b.

( )

The equation of the line perpendicular to the given line through the hiker’s location is y = x + 2.


Find the point of intersection. y = –x + 11 y = x + 2 2 13

132

6.5

y

y

y

=

=

=

+

Substitute y = 6.5 into equation .

6.522

4.5

y xx

x

= += +=

They will reach Hockley Road at (4.5, 6.5). b) Answers may vary. For example: The shortest route might be blocked by fences or thick woods, or it might involve trespassing on private land. Chapter 2 Section 3 Question 25 Page 90 a)


b) Case 1: Connect to cottage A first. Find the equation of the line through points A and T.

2 1

2 1

14 713 6771

y ymx x−

=−−

=

=

=

11 6

y mx bb

b

= +

= +

=

+

=

−

7

( )6 1 13−=

Substitute m = 1 and the coordinates of one point, say (6, 7), to find b.

( )

The equation of the line through points A and T is y = x + 1. The given line has a slope of 1. A line perpendicular to this line has a slope of –1. Substitute m = –1 and the coordinates of the transformer, to find b.

19

y mx bb

b

=

+

The equation of the line perpendicular to the line through points A and T is y = –x + 19. Find the point of intersection. y = x + 1 y = –x + 19 2 20

10yy==

+

Substitute y = 10 into equation .

1011

9

y xx

x

= += +=

The point of intersection is C(9, 10).


Find the length of cable needed.

( ) ( )

( ) ( )

( ) ( )

2 22 1 2 1

2 2

2 2

6 13 4 14

TA

7 10

49 100

14912.2

x x y y= − + −

= − + −

= − + −

= +

=

( ) ( )

( ) ( )

( )

2 22 1 2 1

2 2

22

13 9 6 10

CB

4 4

16 16

325.7

x x y y= − + −

= − + −

= + −

= +

=

The total distance for this route is 12.2 + 5.7, or 17.9 units. Case 2: Connect to cottage B first. Find the equation of the line through B and T. Since this line is vertical, the equation is x = 13. A line perpendicular to this line has a slope of 0. Since it must go through A, the equation is y = 6. The distance from T to B is 14 – 6, or 8 units. The distance from the line through BT to A is 13 – 6, or 7 units. The total distance for this route is 8 + 7, or 15 units. This is the shorter route. Chapter 2 Section 3 Question 26 Page 90 Answers may vary. For example: Using The Geometer’s Sketchpad®: a) Plot the points A, B, and T. Construct line segment AT and the perpendicular from AT to B. Construct point C where the perpendicular meets AT. Then, construct line segment BT and the perpendicular from BT to A. Construct point D where the perpendicular meets BT. b) Measure the lengths of AT, BC, BT, and AD. Use these measurements to show that BT + AD is less than AT + BC. Using Cabri® Jr.: a) Choose Point from the F2 menu, and plot the points A, B, and T. Choose Coord. & Eq. from the F5 menu, and display the coordinates of the points. Adjust the points if necessary. Choose Segment from the F2 menu, and construct line segments AT and BT. Choose Perp. from the F3 menu, and construct the perpendicular from B to AT. Label the point of intersection C. Similarly, construct the perpendicular from A to BT, and label the point of intersection D. b) Choose Measure/D. & Length from the F5 menu, and select AT, BC, BT, and AD. Use these measurements to show that BT + AD is less than AT + BC.


Chapter 2 Section 3 Question 27 Page 91 Solutions for Achievements Checks are shown in the Teacher Resource. Chapter 2 Section 3 Question 28 Page 91 a) b) Find the equations of the three medians. Find the coordinates of the midpoint of side BC.

( )

( ) ( )

( )

1 2 1 2

4

, ,2 2

,2 2

1, 3

x x y yx y + +⎛ ⎞= ⎜ ⎟⎝ ⎠⎛ ⎞+ +

= ⎜ ⎟⎝ ⎠

= −

2 1 5− −−

Find the slope of the median from vertex A.

2 1

2 1

42

3 11

41

y ymx x−

=−

=

−=−

=

4 21 87

y mx bb

bb

= +

= +− =

− −−

1= +

Substitute m = 4 and the coordinates of one endpoint, say (2, 1), to find b.

( )

The equation of the median from vertex A is 4 7y x= − .


Find the coordinates of the midpoint of side AC.

( )

( ) ( )

( )

1 2 1 2, ,2 2

,2 2

0,

1 5

2

2 2

x x y yx y + +⎛ ⎞= ⎜ ⎟⎝ ⎠⎛ ⎞+ +

= ⎜ ⎟⎝ ⎠

=

−−

− Find the slope of the median from vertex B.

( )

2 1

2 1

4

14

1

y ymx x−

=−

=−

−=−

=

2 10 4−− −

Substitute m = 14

and the coordinates of one endpoint, say (0, –2), to find b.

( )124

−

2

0

y mx b

b

b

= +

= +

− =

The equation of the median from vertex B is 1 24

y x= − .

Find the coordinates of the midpoint of side AB.

( )

( )

( )

1 2 1 2

1 1

, ,2 2

,2 2

3,0

4 2

x x y yx y + +⎛ ⎞= ⎜ ⎟⎝ ⎠⎛ ⎞++

= ⎜ ⎟⎝ ⎠

=

−


Find the slope of the median from vertex C.

( )( )

2 1

2 1

551

y ym

0 53 2

x x−

=−

=

=

=

1 33

y mx bb

b

= +

− =

−−− −

0 = +

Substitute m = 1 and the coordinates of one of the endpoints, say (3, 0), to find b.

( )

The equation of the median from vertex C is 3y x= − . Find the point of intersection of the medians from vertices A and B. y = 4x – 7

y = 14

x – 2

Substitute 4x – 7 for y in equation . 1 241 2 Multiply by 4.4

16 28

4 7x = −−

815 20

43

y x

x

x xx

x

= −

− = −=

=

Substitute x = 43

into equation .

4 7

4 7

16 213 353

43

y x= −

⎛ ⎞= −⎜ ⎟⎝ ⎠

= −

= −


⎛ ⎞−⎜ ⎟⎝ ⎠

.


Try these coordinates of the point of intersection in the equation of the median from vertex C.

53

−

y=

=

L.S.

3

3

4 93 3

53

4x

3

= −

= −

= −

= −

R.S.

L.S. = R.S.

All three medians intersect at 4 5,3 3

⎛ −⎜⎝ ⎠

⎞⎟ . This is the centroid of the triangle.

Chapter 2 Section 3 Question 29 Page 91 The median to the hypotenuse of a right triangle is half as long as the hypotenuse. Methods may vary. Using The Geometer’s Sketchpad®: Construct any line and a perpendicular to it. Construct point A where the perpendicular meets the line. Construct point B anywhere on the line and point C anywhere on the perpendicular. Construct line segment BC and the midpoint, D, of BC. Construct line segment AD. Measure and compare the lengths of AD and BC. Observe the ratio of these lengths while dragging point B along the line and point C along the perpendicular. Using Cabri® Jr.: Choose Line from the F2 menu, and construct any line. Choose Perp. from the F3 menu, and construct a line perpendicular to the first line. Choose Point/Intersection from the F2 menu, and construct point A, the intersection of the two lines. Choose Point/Point on, and construct point B on the first line and point C on the second line. Choose Segment from the F2 menu, and construct line segment BC. Choose Midpoint from the F3 menu, and construct point D, the midpoint of BC. Construct line segment AD. Choose Measure/D. & Length from the F5 menu, and select BC and AD. Move the cursor to point B, and press ALPHA. Observe the ratio of the lengths of BC and AD while sliding point B along the first line. Slide point C along the other line.



a) ( ) ( ) ( )

( ) ( ) ( )

2 22 1 2 1 2 1

2 2 2

16 9 1

6 2 3 5 1

6

4

6

1

d x x y y z z= − + − + −

= − + − + −

= + +

=

2

)

b) ( ) ( ) (2 22 1 2 1 2 1d x x 2y y z z= − + − + −

Chapter 2 Section 3 Question 31 Page 91 a) Find the slope of the sewer line and the proposed pipe line for each home.

2 1AB

2 1

60 2080 20406023

y ymx x−

=−−

=

=

=

−

2 1CE

2 1

302032

y ym

40 7050 30

x x−

=−−

=−

−=

= −

2 1DF

2 1

302032

y ym − =x x−50 2065 85

−=

−

=−

= −

CE and DF are both perpendicular to AB. The developer has found the shortest route from each house to the sewer line.

b) ( ) ( )

( ) ( )

( ) ( )

2CE 2 1 2 1

2 2

2 2

7

20 30

400 900

13003

50 30 40 70

6

d x x y y= − + −

= − + −

= + −

= +

=

2 ( ) ( )

( ) ( )

( ) ( )

2 2DF 2 1 2 1

2 2

2 2

7

20 30

400 900

1

6

30036

5 85 50 20

d x x y y= − + −

= − + −

= − +

= +

=

Using the given scale, the length of pipe needed is 2(36 + 36), or 144 m.


c) From part a), the slope of the line through A and B is 23

.

Substitute m = 23

and the coordinates of one point, say (20, 20), to find b.

( )2

4

203

20

0203

203

y mx b

b

b

b

= +

= +

= +

=

The equation of the line through A and B is 2 23 3

y x= +0 .

Find the slope of the line through C and D.

2 1

2 1

20 7085 30

5055

1101

y ymx x−

=−−

=−

−=

= −

Substitute m = 1011

− and the coordinates of one point, say (30, 70), to find b.

( )3007

107011

− 30

011

107011

y mx b

b

b

b

= +

= +

= − +

=

The equation of the line through C and D is 10 107011 11

y x= − + .


Find the point of intersection. 2 23 3

y x= +0

10 107011 11

y x= − +

Substitute 10 107011 11

x− + for y in equation .

2 203 32 20 Multiply by 33.3 3

30 321

10 107011 11

x− +

0 22 22052 2990

57.5

y x

x

x xxx

= +

= +

− + = +− = −

=Substitute x = 57.5 into equation .

( )

2 203 32 23

57.53

45

y x= +

= +

=

0

AB and CD intersect at (57.5, 45). d) Find the length of pipe needed to connect the two houses.

( ) ( )

( ) ( )

( ) ( )

2 22 1 2 1

2 2

2 2

85 30 20 70

CD

55 50

3025 2500

552574

x x y y= − + −

= − + −

= + −

= +

=

CD > CE + DF This route will require more pipe, and should not be used.



! 3!5!7!3 628 800

x ==

Use guess and check to determine that x = 10. Answer A Chapter 2 Section 3 Question 33 Page 91 The length of a side of the smaller square is 24 cm. The area of the smaller square is 576 cm2. The area of the larger square is 676 cm2. The length of a side of the larger square is 26 cm. The perimeter of the larger square is 104 cm. Answer C


Chapter 2 Section 4 Equation for a Circle Chapter 2 Section 4 Question 1 Page 96 a) 2 2 9x y+ = b) 2 2 64x y+ = c) 2 2 100x y+ =


d) 2 2 5x y+ = e) 2 2 12x y+ = f) 2 2 110x y+ =



Chapter 2 Section 4 Question 2 Page 97 a) The radius is 6. Points on the circle include (6, 0), (0, 6), (–6, 0), and (0, –6). b) The radius is 12. Points on the circle include (12, 0), (0, 12), (–12, 0), and (0, –12). c) The radius is 20 . Points on the circle include (2, 4), (–2, 4), (–2, –4), (2, –4), and (4, 2). d) The radius is 50 . Points on the circle include (5, 5), (5, –5), (–5, –5), and (–5, 5). e) The radius is 1.3. Points on the circle include (1.3, 0), (0, 1.3), (–1.3, 0), and (0, –1.3). Chapter 2 Section 4 Question 3 Page 97 a)

( )

2 2 2

22 2

2

2

16 9255

4 3−

r x y

r

rrr

= +

= +

= +

==

An equation for the circle is x2 + y 2 = 25.


b) 2 2

2 2 2

2

2

25 429

5 2

29

r x yrrr

r

= +

= +

= +

=

=

2

An equation for the circle is x2 + y 2 = 29. c)

(

) ( )

2 2 2

2

2

2

9 3645

45

r x y

r

rr

r

= +

=

= +

=

=

2 23 6+− −



d)

( )

2 2 2

27 12− +2 2

2

2

49 144193

193

r x y

r

rr

r

= +

=

=

= +

=



Chapter 2 Section 4 Question 4 Page 97 The radius of the given circle is 34 .

a)

( )

2 21

225

25 9

34

3

r x y= +

= +

= +

=

−

Since r1 = 34 , (5, –3) lies on the circle. b) 2 2

1

2 24 4

16 16

32

r x y= +

= +

= +

=

Since r1 < 34 , (4, 4) lies inside the circle. c)

( )

2 21

2 26 0+−

366

r x y= +

=

=

=

Since r1 > 34 , (–6, 0) lies outside the circle. d)

(

) ( )

2 21

9 25

34

r x y= +

=

= +

=

2 23 5+− −

Since r1 = 34 , (–3, –5) lies on the circle. e)

( )

2 21

222

4 36

40

6

r x y= +

= +

= +

=

−

Since r1 > 34 , (2, –6) lies outside the circle.


f)

( )

2 21

2 23

34

4 0

r x y= +

= +

=

Since r1 = 34 , ( 34 , 0) lies on the circle. Chapter 2 Section 4 Question 5 Page 97 The radius of the given circular path is 1.2 × 104.

2 21

2 2

4

8000 98001.27 10

r x y= +

= +

×

Since r1 > 1.2 × 104, the second satellite is outside the orbit of the first satellite. Chapter 2 Section 4 Question 6 Page 97 Find the diameter.

( ) ( )

( )( ) ( )

2 22 1 2 1

2+

2

2 2

AB

8 6

64 36

1

4 4 3

0010

3

x x y y

− −

= − + −

= − −

= +

= +

==

The diameter is 10 units. The radius is 5 units. The equation of the circle is . 2 2 25x y+ =


Chapter 2 Section 4 Question 7 Page 97 a) 2 2

2 2a2

1001 0

66

8 03

x y

aa

+ =

+ =

== ±

Substituting the coordinates into the equation gives a2 = 36. Therefore, a can be either 6 or –6. b)


Chapter 2 Section 4 Question 8 Page 97 a) Substitute r = 8 into the formula for the circumference of a circle.

( )2250.

83

C r ππ

=

=

20

Approximately 50.3 m of fencing is required. b) Substitute r = 8 into the area formula for a circle.

( )

2

2

1

A rπ

π

=

= 8

The area of the corral is about 201 m2.


Chapter 2 Section 4 Question 9 Page 97 a) b)

( )

2 21

2 2

64 36

100

8 6

10

r x y= +

= +

= +

=

2 21

2 2

36

6 8

64

10010

r x y= +

= +

= +

==

−

=

The points P(–8, 6) and Q(6, 8) both lie on the circle. c) Find the midpoint of chord PQ. Find the slope of chord PQ.

( )

( )

1 2 1 2

8 6 6

, ,2 2

,2 2

1,7

8

x x y yx y + +⎛= ⎜⎝ ⎠

+ +⎛ ⎞= ⎜ ⎟⎝ ⎠

= −

⎞⎟ 2 1

PQ2 1

21417

y ym − =x x8 66 8

−

−−+

7 = +

=

=

=

7 1y mx b

bb− −

= +

=

The slope of the right bisector of PQ must be –7. Substitute m = –7 and the coordinates of the midpoint to find b.

( )0

The equation of the right bisector of PQ is y = –7x. d) The coordinates (0, 0) satisfy the equation y = –7x.


e) Answers may vary. For example: Since the endpoints of any chord lie on a circle, they are equidistant from the centre of the circle. All points equidistant from the endpoints of a line segment lie on the right bisector of the line segment. Therefore, the right bisector of any chord of a circle passes through the centre of the circle.



( )

2 21

2 26 2

36 4

40

r x y= +

= +

= +

=

( )

2 21

222

4 36

40

6

r x y= +

= +

= +

=

− −

The points R(–6, 2) and S(2, –6) both lie on the circle. RS is a chord. c) Find the coordinates of the midpoint, M, of chord RS. Find the slope of OM.

( )

( )

( )

1 2 1 2, ,2 2

,2 2

2,

2 66

2

2

x x y yx y + +⎛= ⎜⎝ ⎠⎛ ⎞++

= ⎜ ⎟⎝ ⎠

= −

⎞⎟

−

2 1OM

2 1

−− 2 02 0

1

y y−m =x x−−−

−=

−=

1 00

y mx bb

by x

= +

==

Substitute m = 1 and coordinates of the origin to find b.

( )0 = +

The equation of the line from the origin to the midpoint of RS is y = x.

d)

( )

2 1RS

2 1

1

y ym

6 22 6

x x−

=−

=

= −

−−−

−

The slopes are negative reciprocals. The line is perpendicular to the chord.



( )

2 21

2 2

16 25

41

4 5

r x y= +

= +

= +

=

−

( ) ( )

2 21

2 2

25 16

41

5 4

r x y= +

= +

= +

=

− −

The points U(–4, 5) and V(–5, –4) both lie on the circle. UV is a chord. c) Find the slope of chord UV.

( )

2 1UV

2 1

4 545

− −−− −

91

9

y ymx x−

=−

=

−=−

=

The slope of a line perpendicular to the chord is 19

− .

Substitute m = 19

− and the coordinates of the origin to find b.

( )109

= +0

0

y mx b

b

b

−

= +

=

The equation of the line from the origin perpendicular to the chord UV is 19

y x= − .


d) Find the coordinates of the midpoint of chord UV.

( )

( ) ( )

1 2 1 2, ,2 2

,2 2

9 1,

4 5 5 4

2 2

x x y yx y

− −

+ +⎛ ⎞= ⎜ ⎟⎝ ⎠⎛ ⎞+ +

= ⎜ ⎟⎝ ⎠⎛=

−

⎞−⎜ ⎟⎝ ⎠

Check that these coordinates satisfy the equation of the line.

1919

92

12

x= −

⎛ ⎞= − ⎜ ⎟⎝ ⎠

12

y=

=

L.S.

=

R.S.

−

L.S. = R.S. The coordinates of the midpoint satisfy the equation. The line passes through the midpoint of the chord. Chapter 2 Section 4 Question 12 Page 98 The right bisector of any chord of a circle passes through the centre of the circle. Methods may vary. Using The Geometer’s Sketchpad®: Construct any circle and a line segment between two points on the circle. Construct the right bisector of the line segment. Choose Animate Point from the Display menu, and animate either endpoint of the line segment. Observe whether the right bisector continues to pass through the centre of the circle. Also, try varying the radius of the circle. Using Cabri® Jr.: Choose Circle from the F2 menu, and construct any circle. Choose Segment from the F2 menu, and construct any line segment with both endpoints on the circumference of the circle. Choose Perp. Bis. from the F3 menu, and select the line segment. Move the cursor to either endpoint of the line segment, and press ALPHA. Drag the endpoint around the circumference of the circle and observe whether the right bisector continues to pass through the centre of the circle. Also, try varying the radius of the circle.


Chapter 2 Section 4 Question 13 Page 98 a), c), d) b)

( ) ( )

2 21

2 2

9 16

25

3 4

5

r x y

− −

= +

= +

= +

==

Point A lies on the circle.


e) Find the slope of OA. 2 1

OA2 1

4 03 0

43

y ymx x−

−

=−

=−

=

−−

The slope of the tangent is 34

− .

Substitute m = 34

− and the coordinates of point A to find b.

( )34 34

= +− −−

944

254

y mx b

b

b

b

= +

− = +

− =

The equation of the tangent is 3 24 4

y x= − −5 .

f) Answers may vary. For example: The tangent touches the circle at point A. Since the circle curves away from the tangent on both sides of point A, the tangent does not touch the circle at any other point. Chapter 2 Section 4 Question 14 Page 98 Answers may vary. For example: The point that is equidistant from the three homes is the centre of the circle that passes through all three homes. A line segment joining any two of the homes is a chord of the circle. The point of intersection of the right bisectors of two of these chords is the centre of the circle. Brandon could draw these right bisectors on a city map and then look for a restaurant near the point where they intersect.


Chapter 2 Section 4 Question 15 Page 98 Determine the height of the tunnel 1.3 m from the centreline.

2 2 2

2.1 3 2 2

2 2

4.54.5

4.5 1.34.31

x yy

yy

+ =

+ =

= −

2

The tunnel is 4.31 m high, and the truck is 4.15 m high. The truck will fit in the tunnel. Chapter 2 Section 4 Question 16 Page 98 Calculate the largest diagonal of the rectangular block.

2 28 912

d = +

This will not fit into the smallest cup, which has a diameter of 10 cm. The blocks will fit into all of the other cups. Chapter 2 Section 4 Question 17 Page 98 a) After 10 s, the ripple has travelled 50 10× , or 500 cm. The equation of the circle is , or . 2 2 500x y+ = 2

2

2 2 250 000x y+ = b) 2 2 2

2 2

2

50 75812590.14

r x yrrr

= +

= +

=

The boat is 90.14 m, or 9014 cm from the anchor.

The ripple will take 901450

, or about 180 s to reach the boat.

c) Answers may vary. For example: Wind or water currents do not move the rowboat or change the shape or speed of the ripple as it travels.


Chapter 2 Section 4 Question 18 Page 98 a) : the region inside the circle centred at (0, 0) with radius 5. 2 2 25x y+ < b) : the region outside the circle centred at (0, 0) with radius 7. 2 2 49x y+ > c) : the region between the circle centred at (0, 0) with radius 5 and the circle centred at (0, 0) with radius 7.

2 225 49x y< + <

Chapter 2 Section 4 Question 19 Page 98 The radius of the small circle is 2. The length of a side of the small square is 2. Calculate the length of the diagonal of the small square. This is the radius of the large circle.

2 22 2

8

d

d

= +

=

2

An equation of the large circle is 2 2 8x y+ = .


Chapter 2 Section 4 Question 20 Page 99 a) The equation of the circle is . 2 2 16x y+ = b) B(0, 2), D(2, 0) c) The y-intercept of the line through A and B is 2. Find the slope.

2 1AB

2 1

0042

2

2

14

y ymx x−

=−−

=−

−=

= −

The equation of the line through A and B is 1 22

y x= − + .

The y-intercept of the line through C and D is 4. Find the slope.

2 1CD

2 1

4

2

20 4

0

2

y ymx x−

=−

=−

−=

= −

−

The equation of the line through C and D is 2 4y x= − + .


d) Find the coordinates of the point of intersection. 1 22

y x= − +

2 4y x= − + Substitute –2x + 4 for y in equation .

1 221 2 Multiply by 2.2

4 8 43 4

2 4x− + = −

43

y x

x

x xx

x

= − +

+

− + = − +− = −

=

Substitute x = 43

into equation .

2 4

2 43

4

8 123 3

43

y x= − +

⎛ ⎞= − +⎜ ⎟⎝ ⎠

= − +

=

The coordinates of point E are 4 4,3 3

⎛ ⎞⎜ ⎟⎝ ⎠

.

e) The area of the quarter circle is ( )21 4 44π π= .

The area of ΔOBA is 1 4 2 42× × = .

The area of ΔBCE is 1 422 3× × =

43

.

The shaded area is 44 4 43 3

π π− − = −16 , or about 7.2 square units.



a) The boat takes 102

, or 5 s to move 10 m. In this time, the waves travel 1 5× , or 5 m.

b)

25

c) At the points of intersection, the waves add together to form a V-shaped wake behind the boat. Chapter 2 Section 4 Question 22 Page 99 The equation of a circle with centre at (4, 3) and radius 5 is ( ) ( )2 24 3x y− + − = . Chapter 2 Section 4 Question 23 Page 99 Answers may vary. For example: No circle with 1 < r < 2 has any points for which both the x- and y-coordinates are integers.



b) From the graph, the equation of the right bisector of QR is 1 42

y x= − + , the equation of the

right bisector of RS is , and the equation of the right bisector of QS is . 1y x= + 3 9y x= − +The circumcentre is at (2, 3). Check that the coordinates satisfy all three equations.

3y=

=L.S.

( )

1 4212

23

4

3

x= − +

= − +

=

.R.S y==

L.S. 11

32x= +R.S.

= +=

3y=

=L.S.

( )3 93 9

32

x= − +

= − +

=

R.S.

L.S. = R.S. L.S. = R.S. L.S. = R.S. The coordinates of the circumcentre satisfy the equations of all three right bisectors.


c) ( ) ( )

( )( ) ( )

2 22 1 2 1

2 2

2 2

2

4 3

55

3

2

2 0

QC x x y y

−

= − + −

= − −

= +

==

)

+

( ) (

( ) ( )

( )

2 22 1 2 1

2 2

22

2 2

RC

25

3

0 5

5

8

x x y y= − + −

= − + −

= + −

==

( ) ( )

( ) ( )

( )

2 22 1 2 1

2 2

2 2

2 7

SC

25

3

5 0

5

3

x x y y= − + −

= − + −

= − +

==

The distance from the circumcentre to each vertex is 5 units. d) The circle has radius 5 and centre (2, 3). e) Answers may vary. For example: Using The Geometer’s Sketchpad®: Construct ΔQRS and the right bisector of each side. Construct the point of intersection of the right bisectors and confirm that all three intersect at the same point. Measure the distance from each vertex to the point of intersection of the right bisectors. The distance in part c) is the radius of the circle. Display the coordinates of the point of intersection, which is the centre of the circle. Using Cabri® Jr.: Choose Triangle from the F2 menu, and construct ΔQRS. Choose Coord. & Eq. from the F5 menu, and check the placement of the vertices. Adjust the vertices if necessary. Choose Perp. Bis. from the F3 menu, and select each side of ΔQRS. Choose Point/Intersection from the F2 menu, and select the three right bisectors. Choose Measure/D. & Length from the F5 menu, and measure the distance from each vertex to the point of intersection of the right bisectors. The distance in part c) is the radius of the circle. Choose Coord. & Eq. from the F5 menu, and select the point of intersection to display the coordinates of the centre of the circle.



2 2

22 2

kx ky rrx yk

+ =

+ =

2

The radius of the circle is 2r r

k k= .

Chapter 2 Section 4 Question 26 Page 99 a) If a < b, the graph is an ellipse (a type of oval) with its length along the x-axis. b) If a > b, the graph is an ellipse with its length along the y-axis.


Chapter 2 Review Chapter 2 Review Question 1 Page 100

a) ( )

( )

1 2 1 2

3 5 2

, ,2 2

,2 2

1,2

6

x x y yx y + +⎛ ⎞= ⎜ ⎟⎝ ⎠

+ +⎛ ⎞⎜

−−= ⎟⎝ ⎠

= The coordinates of the midpoint of AB are (1, 2).

b) ( )

( )

( )

1 2 1 2, ,2 2

,2 2

1.

1 4

,

5

5 2

1

x x y yx y + +⎛ ⎞= ⎜ ⎟⎝ ⎠⎛ ⎞++

= ⎜ ⎟⎝ ⎠

=

−−

The coordinates of the midpoint of CD are (1.5, 2).

c) ( )

( )

1 2 1 2, ,2 2

,2 2

2.

2 7 4 4

5,4

x x y yx y + +⎛ ⎞= ⎜ ⎟⎝ ⎠

+ +⎛ ⎞= ⎜ ⎟⎝ ⎠

=

−

The coordinates of the midpoint of EF are (2.5, 4).


d) ( )

( )

1 2 1 2

20 2

, ,2 2

,2

76 352

7 1

148,

5

2

0

31

x x y yx y + +⎛ ⎞= ⎜ ⎟⎝ ⎠

+ +⎛ ⎞= ⎜ ⎟⎝ ⎠

=

The coordinates of the midpoint of GH are (148, 231). Chapter 2 Review Question 2 Page 100

a) ( )

( ) ( )

( )

1 2 1 2, ,2 2

,2 2

1

3

, 5.5

x x y yx y + +⎛ ⎞= ⎜ ⎟⎝ ⎠⎛ ⎞+

= ⎜ ⎟⎝ ⎠

= − −

5 5 6− − −+

The coordinates of the midpoint of J(3, –8) and K(–5, –6) are (–1, –5.5).

b) ( )

( )

( )

1 2 1 2

8 2

, ,2 2

,2 2

4,3

4 4

x x y yx y + +⎛ ⎞= ⎜ ⎟⎝ ⎠⎛ ⎞++

= ⎜ ⎟⎝ ⎠

=

−

The coordinates of the midpoint of L(4, 8) and N(4, –2) are (4, 3).


Chapter 2 Review Question 3 Page 100 a), c) b) Midpoint of PQ: Midpoint of PR: Midpoint of RQ:

( )

( )

1 2 1 2

2 6 5

, ,2 2

,2 2

2,5

5

x x y yx y + +⎛ ⎞= ⎜ ⎟⎝ ⎠

+ +⎛ ⎞= ⎜ ⎟⎝ ⎠

=

( )

( )

( )

1 2 1 2, ,2 2

,2 2

0,

5 72

1

2

x x y yx y + +⎛ ⎞= ⎜ ⎟⎝ ⎠⎛ ⎞++

= ⎜ ⎟⎝ ⎠

=

( )

( )

1 2 1 2

2 6 7

, ,2 2

,2 2

4, 1

5

x x y yx y + +⎛ ⎞= ⎜ ⎟⎝ ⎠

+ +⎛ ⎞= ⎜ ⎟⎝ ⎠

= −

− −−−

− The coordinates of the midpoint of PQ are (2, 5), the coordinates of the midpoint of PR are (0, –1), and the coordinates of the midpoint of QR are (4, –1).

c) The smaller triangle is similar to ΔPQR, and has 14

the area.


Chapter 2 Review Question 4 Page 100 a) b) Find the coordinates of the midpoint of TV.

( )

( )

( )

1 2 1 2, ,2 2

,2 2

2,

6 48

1

4

x x y yx y + +⎛ ⎞= ⎜ ⎟⎝ ⎠⎛ ⎞++

= ⎜ ⎟⎝ ⎠

= −

−−


( )

2 1

2 1

94

y ym

10 12 2

x x−

=−

=

=

−−−

Substitute m = 94


( )912

112

y mx b

b

b

= +

= − +

=

91 24

b= − +

The equation of the median from vertex U is 9 14 2

y x 1= + .


c) Find the slope of UV. 2 1

2 1

4 104 27

y ymx x−

=−

=

= −

−−

−

The slope of side UV is –7. The slope of the altitude from vertex T is 17

.

Substitute m = 17

and the coordinates of point T to find b.

( )867

507

y mx b

b

b

= +

= − +

=

16 87

b= − +

The equation of the altitude from vertex T is 1 57 7

y x= +0 .


d) Find the coordinates of the midpoint of TU.

( )

( )

1 2 1 2

8 2 6

, ,2 2

,2 2

3,8

10

x x y yx y + +⎛ ⎞= ⎜ ⎟⎝ ⎠

+ +⎛ ⎞= ⎜ ⎟⎝ ⎠

= −

−

Find the slope of TU.

( )

2 1

2 1

10 62 84

1025

y ymx x−

=−−

=−

=

=

−

The slope of TU is 25

. The slope of the right bisector of TU is 52

− .

Substitute m = 25

and the coordinates of the midpoint to find b.

( )58 32

= +−−

1582

12

y mx b

b

b

b

= +

= +

=

The equation of the right bisector of TU is 5 12 2

y x= − + .


Chapter 2 Review Question 5 Page 100 a) b) A line segment joining the midpoints of two sides of a triangle is parallel to the third side. Find the slopes of the sides of ∆EDF.

( )

2 1ED

2 1

1 34 2

2613

y ymx x−

=−

=−

−=

= −

−−

( )

2 1EF

2 1

73

y ym

4 31 2

x x−

=−−−−

−

= −

= ( )

2 1FD

2 1

53

y ym

1 44 1

x x−

−=

−

=

=

−−

Substitute each slope and coordinates of the respective midpoint to find b.

( )14 13

=− −

143

113

y mx b

b

b

b

= +

+

− +

−

= −

=

( )2813

313

y mx b

b

b

= +

= − +

71 43

b= − +

=

( )1033

193

y mx b

b

b

=

3 253

b

+

= − +

= − +

=

The equation of the side through F is 1 13 3

y x 1= − − .

The equation of the side through D is 7 33 3

y x= − +1 .

The equation of the side through E is 5 193 3

y x= + .


Find the point of intersection for each pair of lines. 1 13 3

y x= − −1

7 33 3

y x= − +1

Substitute 1 13 3

x− −1 for y in equation .

7 313 37 31 Multiply by 3.3 3

11

1 113 3

x = −−−

7 316 42

7

y x

x

x xxx

= − +

+

− − = − +==

Substitute x = 7 into equation .

( )

1 113 31 13 37 113 36

1

y x= − −

= − −

= − −

= −

7

The coordinates of vertex R are (7, –6).


1 13 3

y x= − −1

5 193 3

y x= +

Substitute 1 13 3

x− −1 for y in equation .

5 193 35 19 Multiply by 3.

31 113 3

x = +− −3

11 5 196 30

5

y x

x

x xxx

= +

− − = +− =

= −Substitute x = 7 into equation .

( )5

1 113 31 13 3

5 113 32

y x= − −

= − −

−

1

=

= −

−

The coordinates of vertex Q are (–5, –2).


7 33 3

y x= − +1

5 193 3

y x= +

Substitute 7 33 3

x− +1 for y in equation .

5 193 35 19 Multiply by 3.

7 3133 3

x = ++−3

7 31 5 1912 2

11

y x

x

x xxx

= +

− + = +− = −

=Substitute x = 1 into equation .

( )1

7 313 37 33 37 313 3

8

y x= − +

= − +

= − +

=

1

The coordinates of point P are (1, 8). c) Answers may vary. For example: Using The Geometer’s Sketchpad®: Plot points D, E, and F. Construct line segments DE, EF, and DF. Construct a line through D parallel to EF, a line through E parallel to DF, and a line through F parallel to DE. Construct the points of intersection and display their coordinates. Using Cabri® Jr.: Choose Point from the F2 menu, and construct ΔDEF. Choose Coord. & Eq. from the F5 menu, and check the placement of the midpoints. Adjust the placement if necessary. Choose Segment from the F2 menu, and construct line segments DE, EF, and DF. Choose Parallel from the F3 menu, and construct a line through D parallel to EF, a line through E parallel to DF, and a line through F parallel to DE. Choose Point/Intersection from the F2 menu, and construct the three points of intersection of the lines. Choose Coord. & Eq. from the F5 menu, and select the points of intersection.


Chapter 2 Review Question 6 Page 101

a) ( ) ( )

( ) ( )

( )

2 22 1 2 1

2 2

22

8

5 12

16913

3 5 17

d x x y y= − + −

= − + −

= + −

==


b) ( ) ( )

( )) ( )( )(

2 22 1 2 1

2 2

4

6

0

8

1010

d x x y y= − + −

= −

= +

==

2 2

2 3 5+ −− −

The length of line segment CD is 10 units.

c) ( ) ( )

( )( ) ( )( )

2 22 1 2 1

2 2

2 2

5

9 5

3

06

2

1

4

d x x y y

−

= − + −

=

+

−− + −

=

= The length of line segment EF is 106 units.


d)

( ) ( )

( )

( )

2 22 1 2 1

22

22

12 5 1

15 102

6

0

254112

02

2

d x x y y= − + −

⎛ ⎞⎛ ⎞= − + −⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

⎛ ⎞= + −

=

−

⎟⎠

=

−

⎜⎝

The length of line segment GH is 1122

units.



a) ( ) ( )

( ) ( )( )

2 22 1 2 1

2

2 2

4

0 10

10010

4 8

d x x y y= − + −

= − +

= +

==

−

2

2−

The length of line segment JK is 10 units.

b) ( ) ( )

( )( ) ( )( )( )

2 22 1 2 1

2212 5

16913

d x x y y= − + −

= + −

= + −

==

)

2 2

3 15 12 7− −−− −

The length of line segment MN is 13 units.

c) ( ) (

( ) ( )

( ) ( )

2 22 1 2 1

2 28 8

128

d x x y y= − + −

=

= − + −

=

2 23 5 2 6− + −− −

The length of line segment PQ is 128 units.

d) ( ) ( )

( ) ( )( )( )

2 22 1 2 1

2

2 25 6

5

61

d x x y y= − + −

= −

= − +

−+ −

=

2

1 4 1−

The length of line segment RS is 61 units.


e) ( ) ( )

( ) ( )

( )

2 22 1 2 1

2 2

2 2

2 7

9 0

4

81

4

9

d x x y y= − + −

= − + −

=

=

−

− +

= The length of line segment TU is 9 units.

f) ( ) ( )

( )( ) ( )( )

2 22 1 2 1

2 28 1

65

3 5

x y y

−

= − + −

= −

= +

=

d x 2 2

5 6+ −− −

The length of line segment VW is 65 units.


Chapter 2 Review Question 8 Page 101 a) Find the coordinates of the midpoint of side BC.

( )

( )

1 2 1 2, ,2 2

,2 2

5 ,

1 4

3

7 1

2

x x y yx y + +⎛ ⎞= ⎜ ⎟⎝ ⎠⎛ ⎞++

= ⎜ ⎟⎝ ⎠⎛

−

⎞= ⎜ ⎟⎝ ⎠


( ) ( )

( ) ( )( )

2 22 1 2 1

22

2215 5

2

3254

5 5 3 22

d x x y y= − + −

⎛ ⎞= − + −⎜ ⎟⎝ ⎠

⎛ ⎞= +⎜ ⎟⎝ ⎠

=

− −

The length of the median from vertex A is 3254

.


b) Find the lengths of the sides.

( ) ( )

( )( ) ( )( )

2 22 1 2 1

2 2

AB

9 1

82

4 5

x x y y

−

= − + −

= −

= +

=

)

2 2

21+ −− −

( ) (

( ) ( )

( )

2 22 1 2 1

2 2

22

4

C

1

B

3

1

8

3

7

7

x x y y= − + −

= −

= + −

=

+ −−

( ) ( )

( )( ) ( )( )

2 22 1 2 1

2 2

2 2

1− +

AC

6 9

17

7

1

5 2

x x y y

−

= − + −

=

+

−−

=

=

The perimeter of the triangle is 82 73 117+ + , or about 28.4.


Chapter 2 Review Question 9 Page 101 a) b) Check the slopes of the sides of ∆DEF.

( )

2 1ED

2 1

615

25

y ymx x−

=−

=

−=

= −

5 15 10

−−−−

( )

2 1EF

2 1

113

1 110

03−

1

y ymx x−

=−−

( )

2 1FD

2 1

=−

= −

−5 10

5 3−−−

52

y ymx x−

=−

−

=

=

ΔDEF is a right triangle, with the right angle at D. c) Find the length of sides ED and FD.

( ) ( )

( )) ( )(( )

2 22 1 2 1

22

5

ED

15

2

6

61

x x y y= − + −

= −

= + −

=

2 21 50 1+ −− −

( ) ( )

( ) ( )( )

2 22 1 2 1

22

2 2

5 3

F

5 1

D

2 5

0

29

x x y y

− −

= − + −

= − + −

= +

=

( )( )29 126

121

12

2

43

A bh=

=

=

The area of ΔDEF is 1432

square units.


d) Answers may vary. For example: Using The Geometer’s Sketchpad®: Construct ΔDEF. Measure the angles and side lengths. Select the interior of ΔDEF, and choose Area from the Measure menu. Using Cabri® Jr.: Choose Triangle from the F2 menu, and construct ΔDEF. Choose Coord. & Eq. from the F5 menu, and check the placement of the midpoints. Adjust the vertices if necessary. Choose Measure/D. & Length from the F5 menu, and select the sides of ΔDEF. Choose Measure/Angle, and select the angles of ΔDEF. Choose Measure/Area, and select ΔDEF. Chapter 2 Review Question 10 Page 102 Find the lengths of the sides of ΔABC.

( ) ( )

( )( ) ( )

2 22 1 2 1

2 2

2 2

5 2

A

1

2

B

7

5

1

0

x x y y= − + −

= − −

= +

=

)

+−

( ) (

( )( ) ( )

2 22 1 2 1

2 2

2 2

1

AC

3 4

55

5

2

2 1

x x y y

−

= − + −

= − −

= +

==

)

+

( ) (

( ) ( )

( )

2 22 1 2 1

2 2

22

5 1

CB

25

2

4 3

5

5

x x y y= − + −

= − + −

= + −

==

Since AC = CB, ΔABC is isosceles.


Chapter 2 Review Question 11 Page 102 a) Find the slopes of the sides of ΔDEF.

( )

2 1DE

2 1

52

52

y ymx x−

=−

=

−=−

=

2 74 2−−− −

( )

2 1EF

2 1

2 26 4

4

5

102

y ymx x−

=−−

=−

−=

= −

−−

The slopes are negative reciprocals. DE is perpendicular to EF. Therefore, ΔDEF is a right triangle with ∠DEF = 90°.

b) ( )

( )

1 2 1 2, ,2 2

,2 2

52

72 6

,2

2

x x y yx y + +⎛ ⎞= ⎜ ⎟⎝ ⎠⎛ ⎞++

= ⎜ ⎟⎝ ⎠⎛

−−

⎞= ⎜ ⎟⎝ ⎠

The midpoint of the hypotenuse is M 52,2

⎛ ⎞⎜ ⎟⎝ ⎠

.


c) ( ) ( )

( )( )

2 22 1 2 1

22+−

22

52 2 7

DM

145

2

942

4

x x y y= − + −

⎛ ⎞= − −⎜ ⎟⎝ ⎠

⎛ ⎞= + −⎜ ⎟⎝ ⎠

=

)

( ) (

( )( )

2 22 1 2 1

22+−

22

52 4 2

EM

145

2

162

4

x x y y= − + −

⎛ ⎞= − −⎜ ⎟⎝ ⎠

⎛ ⎞= + ⎜ ⎟⎝

=

⎠

)

( ) (

( ) ( )

( )

2 22 1 2 1

22

22

52 6 2

FM

942

1454

2

x x y y= − + −

⎛ ⎞= − + −⎜ ⎟⎝ ⎠

⎛ ⎞= − + ⎜ ⎟⎝ ⎠

=

−

The midpoint of the hypotenuse is equidistance from all three vertices.



a) ( ) ( )

( ) ( )

( )

2 22 1 2 1

2 2

22

65 45 4

20 20

0

80028.

60

3

d x x y y= − + −

= − + −

= + −

=

The distance from A to B is about 28.3 km.

b) ( )

( )

1 2 1 2

45 65

, ,2 2

,2 2

60 4

5

0

55, 0

x x y yx y + +⎛ ⎞= ⎜ ⎟⎝ ⎠

+ +⎛ ⎞= ⎜ ⎟⎝ ⎠

=

c) Find the equation of line representing the branch pipeline.

2 1AB

2 1

40 6065 45

20201

y ymx x−

=−−

=−

−=

= −

1 555

y mx bb

b

= +

= +

− =

The slope of the line representing the pipeline is –1. The slope of the branch pipeline is 1. Substitute m = 1 and the coordinates of the midpoint to find b.

( )50

54

The equation of the line representing the branch line is y = x – 5. Check if the given point satisfies the equation.

y==

L.S.

55

5863x= −

= −=

R.S.

L.S. ≠ R.S. Point C is not on the branch pipeline.


d) Find the equation of the main pipeline. Substitute m = –1 and the coordinates of one point, say ( 45, 60) to find.

( )60 1 45−

54

105

y mx bb

b

= +

+=

=

1 639

y mx bb

b

= +

= +

− =

The equation of the main pipeline is y = –x + 105. Find the equation of the line perpendicular to AB that passes through C. Substitute m = 1 and the coordinates of point C to find b.

( )

The equation of the line perpendicular to AB that passes through C is y = x – 9. Find the intersection with the pipeline. y = –x + 105 y = x – 9 2 96

48yy==

+

Substitute y = 48 into equation .

4899

57

y xxx

= −= −=

The shortest route from C to the pipeline is a straight line that meets the pipeline at (57, 48).



The slope of the given line is 3. The slope of a line perpendicular to this line is 13

− .

The equation of a line perpendicular to the given line that passes through the origin is 13

y x= − .

Find the intersection of the two lines.

3 10y x= − 13

y x= −

Substitute 3x – 10 for y in equation . 131 Multiply by 3.

3 10x = −−

39 30

1 303

0

y x

x

x xxx

= −

− = −==

0

)

Substitute x = 3 into equation .

( )3 103 1

13

y x= −

= −

= −

The point of intersection is (3, –1).

( ) (

( ) ( )

( )

2 22 1 2 1

2 2

22

3 0 1 0

3 1

10

d x x y y= − + −

= − + −

= + −

=

−

The shortest distance from the origin to the given line is 10 .


Chapter 2 Review Question 14 Page 102 a) 2 2 16x y+ = b)

( )

2 2 2

2 2

2

5

34

23

x y r

r

r

+ =

=

+ =−

An equation for the circle is . 2 2 34x y+ = c) 2 2 2

2 2 2

2 2

529.16.4

x y rx yx y

+ =

+ =

+ =


Chapter 2 Review Question 15 Page 103 a) 2 2 2

22 2

2 2

142

814

x y r

x y

x y

+ =

⎛ ⎞+ = ⎜ ⎟⎝ ⎠

+ =

b) 2 2 2

22 2

2 2

142

49

x y r

x y

x y

+ =

⎛ ⎞+ = ⎜ ⎟⎝ ⎠

+ =

c)

( )2 2 2

22 2

2 2

1

2

24

12

x y r

x y

x y

+ =

+ =

+ =

2

2 2

2

765

d) 2 2x y r

rr

+ =

=

+ =

An equation for the circle is . 2 2 65x y+ =


Chapter 2 Review Question 16 Page 103 a) The radius of the given circle is 40 .

(

) ( )

2 21

4 36

40

r x y= +

=

= +

=

2 22 6+− −

Point A lies on the circle. b) The y –intercept of the radius OA is 0. Find the slope of OA.

2 1OA

2 1

6 02 0

3

y ymx x−

=−

=

=

−−

−−

The equation of a line containing the radius OA is y = 3x.

c) The slope of a line perpendicular to OA is 13

− .

Substitute m = 13

− and the coordinates of point A to find b.

( )16 23

= +− −−

263

203

y mx b

b

b

b

= +

− = +

− =

The equation of a line that passes through A and is perpendicular to OA is 1 23 3

y x= − −0 .


d) e) Answers may vary. For example: On either side of point A, the circle curves away from the tangent line. Chapter 2 Review Question 17 Page 103 a) The radius of the given circle is 10 . Verify that the points lie on the circle.

( )

2 21

2 23 1+−

9 1

10

r x y= +

= +

=

=

2 21

2 21 3

1 9

10

r x y= +

= +

= +

=

Both A and B lie on the circle. AB is a chord. b) Find the midpoint of chord AB.

( )

( )

1 2 1 2

3 1 1

, ,2 2

,2 2

1,2

3

x x y yx y + +⎛ ⎞= ⎜ ⎟⎝ ⎠

+ +⎛ ⎞= ⎜ ⎟⎝ ⎠

= −

−

Find the slope of chord AB.

( )

2 1AB

2 1

12

24

y ym

31 3

1x x−

=−

=

=

=

−−−

The slope of chord AB is 12

. The slope of the right bisector is –2.

Substitute m = –2 and the coordinates of the midpoint to find b.


( )2 2 1= +−−

2 20

y mx bb

bb

= +

= +=

The equation of the right bisector is y = –2x. c) The coordinates of the origin satisfy the equation of the right bisector. The right bisector passes through the origin, which is the centre of the circle. Chapter 2 Review Question 18 Page 103 The radius of the signal area is 20 km. Check that the cell phone user is within the circular area.

2 21

2 213 15

169 225

39419.8

r x y= +

= +

= +

=

The user is less than 20 km from the tower. Signals can be received.


Chapter 2 Chapter Test Chapter 2 Chapter Test Question 1 Page 104

( )

( )

1 2 1 2

3 1 3, 5

, ,2 2

2 21,1

x x y yx y + +⎛ ⎞= ⎜ ⎟⎝ ⎠⎛ ⎞= ⎜ ⎟⎝ ⎠

= −

)

+ +− −

Answer C Chapter 2 Chapter Test Question 2 Page 104

( ) (

( )( ) ( )

( )

2 22 1 2 1

24 2+ −−

2

226 6

72

1 5

d x x y y

−

= − + −

= −

= + −

=

Answer C Chapter 2 Chapter Test Question 3 Page 104

2 2 16x y+ = Answer D


Chapter 2 Chapter Test Question 4 Page 104

The coordinates of the midpoint of EF are 13,2

⎛ ⎞− −⎜ ⎟⎝ ⎠

. The

length of EF is 7 units. The coordinates of the midpoint of GH are (1, 4). The length of GH is 4 units. Find the coordinates of the midpoint of IJ and the length of IJ.

( ) 1 2 1 2

2 3

, ,2 2

,2 2

1 5,2

6 1

2

x x y yx y

− −

+ +⎛ ⎞= ⎜ ⎟⎝ ⎠

+ +⎛ ⎞= ⎜ ⎟⎝ ⎠⎛ ⎞= −⎜ ⎟⎝ ⎠

( ) ( )

( )( ) ( )( )

2 22 1 2 1

2 2

3

IJ

5 7

74

2 22 1 6

x x y y= − + −

= −

= +

=

−−−

The coordinates of the midpoint of IJ are 1 5,2 2

⎛ −⎜⎝ ⎠

⎞⎟ . The length of IJ is 74 units.

Find the coordinates of the midpoint of LK and the length of LK.

( )

( )

1 2 1 2

5 26 4

, ,2 2

,2 2

75,2

x x y yx y + +⎛ ⎞= ⎜ ⎟⎝ ⎠⎛ ⎞++

= ⎜ ⎟⎝ ⎠⎛ ⎞= −⎜ ⎟⎝ ⎠

)

− −

( ) (

( ) ( )( )( )

2 22 1 2 1

2

22

6

L

4

K

2 3

13

x2

5 2

x y y= − + −

= −

= + −

=

−−−

The coordinates of the midpoint of KL are 75,2

⎛ ⎞−⎜ . The length of LK is ⎟⎝ ⎠

13 units.


Chapter 2 Chapter Test Question 5 Page 104 a) 2 2 36x y+ = b)

( )

2 2

2

2

2

2

3

18

23

x y r

r

r

−

+ =

=

+ =

An equation for the circle is . 2 2 18x y+ = Chapter 2 Chapter Test Question 6 Page 104 Answers may vary. For example: Rachel is not correct. Any point on the right bisector of BC is equidistant from points B and C. It need not be the midpoint.



a) ( ) ( )

( ) ( )

2 22 1 2 1

2 2

2 2

14 4 8 3

PS

10 5

12511.2

x x y y= − + −

= − + −

= +

=

The schools are about 11.2 km apart.

b) ( )

( )

1 2 1 2, ,2 2

,2 2

9,5.5

4 14 3 8

x x y yx y + +⎛ ⎞= ⎜ ⎟⎝ ⎠

+ +⎛ ⎞= ⎜ ⎟⎝ ⎠

=

The coordinates of Jason's home are (9, 5.5). c) Answers may vary. For example: Any point on the perpendicular bisector of PS will be equidistant from the two schools. d) Find the slope of PS.

2 1PS

2 1

510

3814 4

12

y ymx x−

=−−

=−

=

=

y mx bb

bb

y x

= +

= +

= − +== − +

The slope of the right bisector of PS is –2. Substitute m = –2 and the coordinates of the midpoint to find b.

( )5.5 18

23.52 23.

9

5

5.5 2−

The equation of the right bisector of PS is 2 23.5y x= − + .


Chapter 2 Chapter Test Question 8 Page 104 a)

)

b) ( ) (

( )( ) ( )

( )

2 22 1 2 1

21 2

22

AB

4 2

2

1

0

2 2

x x y y

−

= − + −

= − −

= + −

=

)

+ −

( ) (

( )( ) ( )

2 22 1 2 1

2 2

2 2

0 2

A

4

5

C

2

2

1

0

x x y y= − + −

= − −

= +

=

)

+−

( ) (

( ) ( )( )( )

2 22 1 2 1

2

2 2

0 2 5

BC

2 6

40

x2

1

x y y= − + −

= −

= − +

−+ −

=


c) Check the slopes of the sides of ∆ABC.

( )

2 1AC

2 1

42

25 1

0

2

y ymx x−

=−

=

=

=

−−−

( )

2 1AB

2 1

2412

y ym

1 12 2

x x−

=−

− −

=

−=

= −

− −

AC is perpendicular to AB, and AB = AC. The triangle is an isosceles right triangle.

d)

( )( )

1212

20 20

10

A bh=

=

= The area of ΔABC is 10 square units. e) Answers may vary. For example: Using The Geometer’s Sketchpad®: Construct ΔABC. Measure each side. Compare the lengths of the sides and the measures of the angles. Select the interior of ΔABC, and choose Area from the Measure menu. Using Cabri® Jr.: Choose Triangle from the F2 menu, and construct ΔABC. Choose Coord. & Eq. from the F5 menu, check the placement of the vertices, and adjust them if necessary. Choose Measure/D. & Length from the F5 menu, and select the sides of ΔABC. Compare the lengths of the sides. Choose Measure/Angle, and select the angles of ΔABC. Choose Measure/Area, and select ΔABC.


Chapter 2 Chapter Test Question 9 Page 104 a) b) Find the coordinates of the midpoint, M, of QP.

( )

( )

1 2 1 2

5 3 2

, ,2 2

,2 2

1,3

4

x x y yx y + +⎛ ⎞= ⎜ ⎟⎝ ⎠

+ +⎛ ⎞= ⎜ ⎟⎝ ⎠

= −

−


( )

2 1MR

2 1

7272

y ym

4 31 1

x x−

=−

=−

−=

= −

− −−

Substitute m = 72

− and the coordinates of one endpoint, say (–1, 3), to find b.

( )73 12

= +−−

732

12

y mx b

b

b

b

= +

= +

− =

The equation of the median from vertex R is 7 12 2

y x= − − .


c) Find the slope of side QP.

( )

2 1QP

2 1

14

28

y ym

43 5

2x x−

=−−

=

=

=

2

2

2

5

34

−−

23

The slope of QP is not the negative reciprocal of the slope of RM. The median is not an altitude. Chapter 2 Chapter Test Question 10 Page 105 a) The coordinates of the other endpoint of the diameter are (–3, 5). b) (3, 5) also lies on the circle. c)

( )

2 2

2

x y r

r

r

−

+ =

=

+ =

An equation for the circle is . 2 2 34x y+ = d) Substitute the coordinates (3, 5) into the equation to see if they satisfy the equation.

2 2 34x y+ =

e) Answers may vary. Possible points include (–3, –5), (–5, 3), (5, 3), and (0, 34 ).



a) ( )

( )

1 2 1 2G , ,2 2

2 4 6 10,2 2

3,8

x x y yx y + +⎛ ⎞= ⎜ ⎟⎝ ⎠

+ +⎛ ⎞= ⎜ ⎟⎝ ⎠

=

( )

( )

1 2 1 2

8 4 2 1

H , ,2 2

,2 2

6,4

0

x x y yx y + +⎛ ⎞= ⎜ ⎟⎝ ⎠

+ +⎛ ⎞= ⎜ ⎟⎝ ⎠

=

−

The coordinates are G(3, 8), and H(6, 4). b) Find the slopes of GH and DE.

2 1GH

2 1

4368

3

4

43

y ymx x−

=−−

=−

−=

= −

2 1DE

2 1

2 68 28

643

y ymx x−

=−

=

−=

= −

)

− −−

Since the slopes are the same, GH is parallel to DE.

c) ( ) (

( ) ( )

( )

2 22 1 2 1

2 2

22

6 3

GH

25

4

3 4

5

8

x x y y= − + −

= − + −

= + −

==

) ( ) (

( ) ( )

( )

2 22 1 2 1

2 2

22

DE

6 8

10010

8 2 2 6

x x y y= − + −

= − + −

= + −

==

−

GH is half the length of DE.


Chapter 2 Chapter Test Question 12 Page 105 a) Check the slopes of the sides of ∆UVW.

2UV

2 1

8

1

42

y ym

5 30 4

x x−

=−

=−

−=−

=

−−

( )( )

2 1WV

2 1

24

21

y ymx x−

=−

=−

5 30 4− −−

−

−=

= −

UV is perpendicular to WV. ΔUVW is a right triangle, with the right angle at V. b) Find the midpoint, M, of side UW.

( )

( ) ( )

( )

1 2 1 2

4

, ,2 2

2 2

0,0

x x y yx y + +⎛ ⎞= ⎜ ⎟⎝ ⎠⎛ ⎞+

= ⎜ ⎟⎝ ⎠

=

)

4 3,

3+− −

Find the length of the median and the hypotenuse.

( ) (

( ) ( )

( )

2 22 1 2 1

2 2

22

0 0 5 0

MV

0 5

255

x x y y= − + −

= − + −

= + −

==

( ) ( )

(− ) ( )

( ) ( )

2 22 1 2 1

2 2

UW

8 6

10010

2 24 4 3 3

x x y y= − + −

=

= − + −

==

− + −− −

The length of the median from the right angle is half as long as the hypotenuse.


c) From part b), the radius of the circle is 5. The equation of the circle is . 2 2 25x y+ = d) Answers may vary. For example: Using The Geometer’s Sketchpad®: Construct ΔUVW, and measure each angle. Construct the midpoint, M, of side UW. Construct line segment VM. Measure the length of UW and of VM. Construct the circle with centre M and radius 5. Select the circle and choose Equation from the Measure menu. Using Cabri® Jr.: Choose Triangle from the F2 menu, and construct ΔUVW. Choose Coord. & Eq. from the F5 menu, check the placement of the vertices, and adjust them if necessary. Choose Measure/Angle from the F5 menu, and select the angles of ΔUVW. Choose Midpoint from the F3 menu, and construct the midpoint, M, of side UW. Choose Segment from the F2 menu, and select points V and M. Choose Measure/D. & Length from the F5 menu, and select UW and VM. Choose Circle from the F2 menu, and construct the circle with centre M and radius 5. Choose Coord. & Eq. from the F5 menu and select the circle.


Chapter 2 Chapter Test Question 13 Page 105 a) b) The boundary of the area is described by the equation . 2 2 400x y+ = c)

( )

2 21

2 28 1

64 256

32017

6

.9

r x y= +

= +

=

2 21

2 24 20

16 400

41620.4

r x y= +

= +

= +

=

−

+

=

Since r1 < 20, Arif is inside the circle. Since r1 > 20, Diane is outside the circle. Arif is in range, but Diane is not.

d) ( ) ( )

( ) ( )

( ) ( )

2 22 1 2 1

2 2

2 2

8 4− 16

12 4

16012 6

0

.

2

d x x y y= − + −

= − + −

+

=

= − −

Diane and Arif are about 12.6 km apart. They are within range of each other. Chapter 2 Chapter Test Question 14 Page 105 Solutions for the Achievement Checks are shown in the Teacher Resource.


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