Mathematics

Introduction to Cartesian SystemSession 3

Session Opener

A government program allows people to collect empty milk bottles and exchange them for bottles full of milk. Four empty bottles may be exchanged for one full bottle. How many bottles of milk can a family drink if it has collected 24 empty bottles?

(a) 7 (b) 6 ( c) 9 (d) 8

Session Opener

(a) 7 (b) 6 ( c) 9 (d) 8

24 empty bottles 6 Milk bottles

1 Milk bottles

3 empty bottles

2 empty bottles

4 empty bottles

1 bottle (borrow) 1 Milk bottles

Session Objectives

1. Transformation of one form of line to other Form

2. Intersection of lines

3. Concurrency of three lines

4. Shortest distance of a point from a given line

5. Distance between two parallel lines

6. Position of a point w.r.t a given line

General Form - Equation of Line

ax by c 0

This form can be transformed to other standard forms to simplify problems

General to Slope- Intercept Form

General form of equation of a line ax+by+c = 0

On rearranging, a cy xb b

a coeff. of xslope b coeff of y c cons tant termIntercept b coeff of y

Illustrative Problem :

Find the slope and y-intercept for line given by 2x + 4y +10 = 0.

On rearranging, 4y 2x 10Solution :

2 10y x4 4

1 5y x2 2

1 5Slope ; y axis int ercept2 2

Comparing with

y = mx + c

General to Intercepts Form

On rearranging, x y 1c ca b

General form of equation of a line ax+by+c = 0

cIntercept on x axis a cIntercept on y axis b

Put y = 0 in general equation , Solve for x

Illustrative Problem : Find the x-intercept and y-intercept for line given by 2x + 4y +10 = 0.

Find the area formed by this line and the coordinate axis.

2x 4y 10Solution :

2x 4y 110 10

5x int ercept 5 ; y int ercept 2

x y 1552

1 5 25Area of OAB . 5 .2 2 4

OA

B

General to Normal Form

Normal form : xcos + ysin - p = 0

General form ax+by+c = 0

O

P

OP = p

General to Normal Form

Normal form : xcos + ysin - p = 0a b c

cos sin p ap bpcos ; sinc c

2 2As cos sin 1

2 2cp

a b

General form ax+by+c = 0

2 2 2 22 2

a p b p 1c c

2 2 2 2a bcos ; sin

a b a b

2 2 2 2 2 2a b cx

a b a b a b

c is +ve

Illustrative Problem

Show that the origin is equidistant from the lines 4x+3y+10 = 0, 5x-12y+26 = 0and 7x+24y = 50

Illustrative ProblemShow that the origin is equidistant from the lines 4x+3y+10 = 0, 5x-12y+26 = 0 and 7x+24y = 50

Solution :

4x 3y 10

Transforming the equations to normal form, we have,

5 12x y 2;13 13

7 24x y 225 25

4 3x y 2;5 5

5x 12y 26

7x 24y 50

Length from origin = 2 units

Class Exercise - I (8 Min.)1. Find the slope of line L : 5x+12y–26 = 0.

Convert the given equation in to normal form.Find length of perpendicular and the slope of the perpendicular.

3. The straight line passing through the point of intersection of the lines L1 : x – 3y +1 = 0 and 2x + 5y – 9 = 0 and having infinite slope is(a) y = 1 (b) 3x + y – 1 = 0 (c) x = 4 (d) None of these

2. The area of triangles made by the lines L1 : 2x + 3y – k = 0, L2 : x = 0, L3 : y = 0 is 48 sq. units. The values of k could be (a) 48 (b) 24 (c) 24 (d) None of these

Solution 1-Class Exercise - I Find the slope of line L : 5x+12y–26 = 0. Convert the given equation in to normal form.Find length of perpendicular from origin and the slope of the perpendicular.

Solution : Rearranging the given equation,

5 12 26x y13 13 13

5x +12y = 262 2Divide by 5 12 13

Length of perpendicular from origin = 2 units

5 12x y 213 13

5 12cos , sin

13 13

12tan (Slope of perp.)5

Solution 2 - Class Exercise - I

Solution :

The area of triangles made by the lines L1 : 2x + 3y – k = 0, L2 : x = 0, L3 : y = 0 is 48 sq. units. The values of k could be (a) 48 (b) 24 (c) 24 (d) None of these

Y

C

X

L1

L2

L3

A

B

Y

C

X

L1

L2

L3

A

B

L2 and L3 are Y-axis and X-axis, triangle is as

1Area of ABC AC BC

2

1x yL : 1k k2 3

k3

k21 k k 48 k 24

2 3 2

Solution 3 - Class Exercise - I

Solution :

The straight line passing through the point of intersection of the lines L1 : x – 3y +1 = 0 and 2x + 5y – 9 = 0 and having infinite slope is(a) y = 1 (b) 3x + y – 1 = 0 (c) x = 4 (d) None of these

Point of intersection of L1 and L2 : (2,1)

Slope is infinite, equation of line will be x = k

Equation of required line is x = 2

Intersection of Two Lines

Consider two linesa1x+b1y+c1 = 0 anda2x+b2y+c2 = 0

Let the point of intersection be P (x1, y1) a1x1+b1y1+c1 = 0 a2x1+b2y1+c2 = 0Solving by cross-multiplication,

1 2 2 1 1 2 2 11 2 2 1 1 2 2 1

b c b c c a c aP ,a b a b a b a b

No need to memorise

Point of intersection will lie on both lines.

Solve two simultaneous equations

Illustrative Problem

Find the point of intersection of lines x + 2y = 9 and 2x – y– 3 = 0.

Solution : Solve two simultaneous

equations x+2y = 9 4x–2y = 6

x = 3 , y = 3

Concurrency of Three LinesThree concurrent lines pass through a common point.

Let the three lines beL1 a1x+b1y+c1 = 0,L2 a2x+b2y+c2 = 0, L3 a3x+b3y+c3 = 0

1 2 2 1 1 2 2 11 2 2 1 1 2 2 1

b c b c c a c aP.O. I. ,a b a b a b a b

P.O.I will satisfy L3

1 2 2 1 1 2 2 13 3 3

1 2 2 1 1 2 2 1

b c b c c a c aa b c 0a b a b a b a b

3 1 2 2 1 3 1 2 2 1 3 1 2 2 1a b c b c b c a c a c a b a b 0

Concurrency of Three Lines

1 1 1

2 2 2

3 3 3

a b ca b c 0a b c

L1 a1x+b1y+c1 = 0,L2 a2x+b2y+c2 = 0, L3 a3x+b3y+c3 = 0

3 1 2 2 1 3 1 2 2 1 3 1 2 2 1a b c b c b c a c a c a b a b 0

Illustrative Problem :

1 1 10 1 5 02 3 k

If the three lines x + y – 1 = 0, y = 5 and 2x + 3y = k are concurrent, the value of k is (a) 1 (b) –1 (c) 0 (d) None of these

Solution :

If the lines are concurrent

1 k 15 1 10 1 2 0

k 15 10 2 0 k 7

Family of Lines Passing through Intersection of two lines

Family of lines passing through point of intersection of L1 and L2 is :

Let the to intersecting lines beL1 a1x+b1y+c1 = 0,L2 a2x+b2y+c2 = 0,

L1+k L2 = 0 Why ?

Problem from booklet

Concurrency of Three LinesAlternate condition for concurrency of three lines :

Iff there exist three non-zero constants p, q, r such thatpL1+qL2+rL3 = 0.

Let the three lines beL1 a1x+b1y+c1 = 0,L2 a2x+b2y+c2 = 0, L3 a3x+b3y+c3 = 0

Illustrative ProblemIf the three lines x + y – 1 = 0, y = 5 and 2x + 3y = k are concurrent, the value of k is

(a) 1 (b) –1 (c) 7 (d) None of these

Solution : p(x+y-1) + q(y-5) + r(2x+3y-k) = 0 x(p+2r) + y(p+q+3r) - (p+5q+rk) = 0

p = -2r , p+q = -3r , p+5q = - r k

p = -2r , q = -r -7r = -rk

7 = k

Class Exercise - II ( 5 Min.)

2. Prove that all lines represented by the equation: (2cosA+3sinA)x + (3cosA – 5sinA )y = ( 5 cosA – 2 sinA) pass through a fixed point for all values of A . Find the coordinates of fixed point.

1. The point through which the family of lines (a + 2b)x + (a + 3b)y – (a + b) = 0

passes for all values of a and b is

(a) (2, –1) (b) (–1, 2) (c) (–2, 1) (d) (2, 1)

Solution 1 - Class Exercise - II The point through which the family of lines (a + 2b)x + (a + 3b)y – (a + b) = 0 passes for all values of a and b is(a) (2,–1) (b) (–1,2) (c) (–2,1) (d) (2,1)

Solution :

The above lines passe through the p.o.i. of the lines

a(x + y – 1) + b(2x + 3y – 1) = 0

Rearranging the given equation

Solving (i) and (ii), we get x = 2 and y = –1

x + y – 1 = 0 ...(i) and 2x + 3y – 1 = 0 ...(ii)

Solution 2 - Class Exercise - II

Solution :

Solving (i) and (ii), we get x = 1, y = 1

Prove that all lines represented by the equation: (2cosA+3sinA)x + (3cosA – 5sinA )y = ( 5 cosA – 2 sinA) pass through a fixed point for all values of A . Find the coordinates of fixed point.

Rearranging the given equation cosA(2x + 3y – 5) + sinA(3x – 5y + 2) = 0

The above lines passe through the p.o.i. of the lines

2x + 3y – 5 = 0...(i) and 3x – 5y + 2 = 0 ...(ii)

Distance of a point from a lineConsider line L, xcos+ysin = p

P (x1, y1)

A

B

O

L

C

and point P (x1, y1).

D

Q

L1

Draw a Line L1 passing through P but parallel to LL1 : xcos+ysin = p1

PQ = CD = OD – OC = p1 - p

p1= x1cos+y1sin

PQ = x1cos+y1sin - p

Distance of a point from a lineConsider line L, xcos+ysin = pand point P (x1, y1).

P (x1, y1)

A

B

OL

CD

QL1

PQ = x1cos+y1sin - p

2 2 2 2a bcos ; sin

a b a b

2 2cp

a b

1 12 2 2 2 2 2

a b cPQ x ya b a b a b

1 12 2

ax by cPQ

a b

Distance of a point from a line

Step I : Convert given equation to ax+by+c = 0

Note : Absolute value

Algorithm

Step II : Put cordinates of the given point (x1,y1) in place of x and y.

ax1+by1+c

Step III : 1 12 2

ax by c

a b

Problem from booklet

Distance between Two Parallel Lines

Y

A

B

O

M

X

L1

N

L2Y

A

B

O

M

X

L1

N

L2

L1 : a1 x + b1 y + c1 = 0 L2 : a2 x + b2 y + c2 = 0

Distance between L1 and L 2

= OM – ON

2 12 2 2 2

2 2 1 1

c cMN

a b a b

Make coefficient of x as +ve.

Illustrative Problem

Find the distance between the parallel lines 3x-4y+9 = 0 and 6x-8y-15 = 0

Solution :

2 2 2 2

9 15Dis tance3 4 6 8

9 15Dis tance 3.3 units5 10

Position of points relative to a lineLet the given line be L ax+by+c = 0Let the given points be P1 (x1, y1)and P2 (x2, y2)Let a point Q (x3, y3) divide line segment P1P2 in ratio m:n

2 1 2 1mx nx my nyQ ,m n m n

If Q lies on L,

2 1 2 1mx nx my nya b c 0m n m n

Positions of Points Relative to a Line

2 2 1 1m ax by c n ax by c 0

1 12 2

ax by cmn ax by c

Ratio positive ax1+by1+c and ax2+by2+c have opposite signs

points are on opposite sides of the line

P1

P2L

Q

Positions of Points Relative to a Line

2 2 1 1m ax by c n ax by c 0

1 12 2

ax by cmn ax by c

Ratio negative ax1+by1+c and

ax2+by2+c have same sign

points are on same side of the line

L

P1

P2Q

Positions of Points Relative to a LineAlgorithm

Step I : Write the line in general formax+by+c = 0

Step II : Determine sign of E1 = ax1+by1+c and E2= ax2+by2+c

If E1 and E2 are of same sign, the point lie on the same side of the line

If E1 and E2 are of opposite sign, the point lie on the opposite sides of the line

Illustrative ProblemAre the points (3, -4) and (2, 6) on the same or opposite sides of the line 3x-4y=8?

Given equation in general form, 3x–4y–8 = 0

Solution :

E1 = 3(3) – 4(–4)–8 = 14

E2 = 3(2) – 4(6)–8 = -26

Opposite side

Class Exercise - III ( 8 Min.)

2. Check whether the origin and (2, 3) lies on the same side of 2x + 5y = 10 or not.

3. If p and p1 are the two perpendiculars from origin on the line: x sec + y cosec = a and x cos - y sin = a cos 2 respectively. Prove that 4p2 + p1

2 = a2.

1. Find the distance between lines y = mx + c and 2y = 2mx + d

Solution 1 -Class Exercise - III Find the distance between lines y = mx + c and 2y = 2mx + d

Solution :The equation of lines can be written as

Distance between parallel lines =

mx – y + c = 0 ...(i) and mx – y + d/2 = 0 ...(ii)

2

dc –2

1 m

Solution 2 -Class Exercise - III Check whether the origin and (2, 3) lies on the same side of 2x + 5y = 10 or not.

Solution :

E(0, 0) = 2(0) +5(0) – 10 = –10

Equation of line is 2x + 5y – 10 = 0

E(2, 3) = 2(2) +5(3) - 10 = 9

Opposite Side

Solution 3 -Class Exercise - III If p and p1 are the two perpendiculars from origin on the line: x sec + y cosec = a and x cos - y sin = a cos 2 respectively. Prove that 4p2 + p1

2 = a2.

Solution :

2 2

–a ap asin cos sin22sec cosec

1 2 2

–acos2p a cos2

sin cos

2 21LHS 4p p

22 2 2a4 sin 2 a cos 2

4 = 4 a2

Thank you