Straight line

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Higher Outcome 1

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Higher Unit 1Higher Unit 1

Distance FormulaThe Midpoint FormulaGradients CollinearityGradients of Perpendicular Lines The Equation of a Straight LineMedian, Altitude & Perpendicular BisectorExam Type Questions

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Higher Outcome 1

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Starter QuestionsStarter Questions

1. Calculate the length of the length AC.

(a)( 1, 2) and ( 5, 10) (b) ( -4, -10) and ( -2,-6)

2. Calculate the coordinates that are halfway between.

6 m

8 m

A

CB

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Higher Outcome 1

Distance FormulaDistance FormulaLength of a straight lineLength of a straight line

A(x1,y1)

B(x2,y2)

x2 – x1

y2 – y1

C

x

y

O

This is just

Pythagoras’ Theorem

2 2 2AB =AC +BC

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Higher Outcome 1

Distance FormulaDistance Formula

The length (distance ) of ANY line can be given by the formula :

2 1 2 1tan ( ) ( )dis ceAB y y x x

Just Pythagoras Theorem in

disguise

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Higher Outcome 1

Finding Mid-Point of a lineFinding Mid-Point of a line

A(x1,y1)

B(x2,y2)

x

y

O

1 21 2 , ,2 2

y yx xM

x1 x2

M

y1

y2

The mid-point between

2 points is given by

Simply add both x coordinates together

and divide by 2.

Then do the same with the y

coordinates.

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Higher Outcome 1

Straight line FactsStraight line Facts

Y – axis Intercept

2 1

2 1

y - yGradient =

x - x

y = mx + c

Another version of the straight line general formula is:

ax + by + c = 0

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Higher Outcome 1

Apr 12, 2023www.mathsrevision.com 7

m < 0

m > 0

m = 0

x = a

y = c

Sloping left to right up has +ve gradient

Sloping left to right down has -ve gradient

Horizontal line has zero gradient.

Vertical line has undefined gradient.

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Higher Outcome 1


m = tan θ

m > 0

Lines with the same gradient

means lines are Parallel

The gradient of a line is ALWAYS

equal to the tangent of the angle

made with the line and the positive x-axisθ

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Higher Outcome 1

CollinearityCollinearity

A

C

x

y

O x1 x2

B

Points are said to be collinear

if they lie on the same straight.

The coordinates A,B C are collinear since they lie on

the same straight line.

D,E,F are not collinear they do not lie on the

same straight line.

D

EF

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Higher Outcome 1

Gradient of perpendicular linesGradient of perpendicular lines

If 2 lines with gradients m1 and m2 are perpendicular then m1 × m2

= -1

When rotated through 90º about the origin A (a, b) → B (-b, a)

-aB(-b,a)

-b

A(a,b)

aO

y

x

- 0

- 0 OA

b bm

a a- 0

-- - 0

OB

a am

b b

- -1-

OA OB

b a abm m

a b ab

Conversely:

If m1 × m2 = -1 then the two lines with gradients m1 and m2 are perpendicular.

-b

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Higher Outcome 1

=

The Equation of the Straight LineThe Equation of the Straight Liney – b = m (x - a)

The equation of any line can be found if we know

the gradient and one point on the line.

O

y

x

P (x, y)

m

A (a, b)y - by - b

x – a

m =y - b

(x – a)m

Gradient,

mPoint (a,

b)

y – b = m ( x – a ) Point on the line ( a, b )

a x

y

b

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Higher Outcome 1

Find the equation of the line which passes through the point

(-1, 3) and is perpendicular to the line with equation 4 1 0x y

Find gradient of given line: 4 1 0 4 1 4x y y x m

Find gradient of perpendicular:1

(using formula 1)1 24 m mm

Find equation:

4 13 0y x

Typical Exam Typical Exam QuestionsQuestions

y – b = m(x – a)

y – 3 = ¼ (x –(-1))

4y – 12 = x + 1

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Higher Outcome 1

Find the equation of the straight line which is parallel to the

line with equation

and which passes through the point (2, –1). Find gradient of given line:

Gradient of parallel line is same:

Find equation:

3 2 1y x

2 3 5x y

2 2

3 33 2 5 5y x y x m


y – b = m(x – a)

y – (-1) = -2/3 (x – 2)

3y + 3 = -2x + 4

M = -2/3

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Higher Outcome 1

A

B C

D


AMedian means a line from vertex

to midpoint of the base.

Altitude means a perpendicular line

from a vertex to the base.

B D C

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Higher Outcome 1


A

B D C

Perpendicular bisector means a line from the vertex

that cuts the base in half and at right angles.

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Higher Outcome 1

Find gradient of the line:

1tan

3

Use table of exact values1 1

tan 303

2 ( 1) 3 1

3 3 0 3 3 3m

tanm

Find the size of the angle a° that the line

joining the points A(0, -1) and B(33, 2)

makes with the positive direction of the x-

axis.

Exam Type QuestionsExam Type Questions

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Higher Outcome 1

A and B are the points (–3, –1) and (5, 5).

Find the equation of

a) the line AB.

b) the perpendicular bisector of AB Find gradient of the AB: 4 3 5y x

Find mid-point of AB

3

4m Find equation of AB

Gradient of AB (perp):4

3m

Use gradient and mid-point to obtain perpendicular bisector AB

Exam Type QuestionsExam Type Questions

(1,2)

4x + 3y = 10

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Higher Outcome 1

The line AB makes an angle of radians

with

the y-axis, as shown in the diagram.

Find the exact value of the gradient of AB.

Find angle between AB and x-axis:2 3 6

Use table of exact values

3

tanm tan6

m

1

3m

(x and y axes are perpendicular.)


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Higher Outcome 1

A triangle ABC has vertices A(4, 3), B(6, 1)

and C(–2, –3) as shown in the diagram.

Find the equation of AM, the median from A.

Find mid-point of BC: 2 1 2 1-(2, 1) Using M ,

2 2

x x y y

Find equation of median AM

Find gradient of median AM

2 1

2 1

-2 Using

-

y ym m

x x

2 5 Using - ( - ) y x y b m x a


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Higher Outcome 1

P(–4, 5), Q(–2, –2) and R(4, 1) are the vertices

of triangle PQR as shown in the diagram.

Find the equation of PS, the altitude from P.

Find gradient of QR:2 1

2 1

-1 Using

2 -

y ym m

x x

Find equation of altitude PS

Find gradient of PS (perpendicular to QR)

1 22 ( 1) m m m

2 3 0 Using ( ) y x y b m x a


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Higher Outcome 1

The lines and

make angles of a and b with the positive direction of the x-axis, as shown in the diagram.

a) Find the values of a and b

b) Hence find the acute angle between the two given lines.

2m

Find supplement of b 180 135 45

2 4y x 13x y

2 4y x

13x y 1m

Find a° tan 2 63a a

Find b° tan 1 135b b

angle between two lines

Use angle sum triangle = 180°

72°


45o

72o

63o 135o

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Higher Outcome 1Triangle ABC has vertices

A(–1, 6), B(–3, –2) and C(5, 2)

Find:

a) the equation of the line p, the median from C of triangle ABC.

b) the equation of the line q, the perpendicular bisector of BC.

c) the co-ordinates of the point of intersection of the lines p and q.

Find mid-point of AB

Find equation of p

2y

Find gradient of p(-2, 2)

Find mid-point of BC

(1, 0) Find gradient of BC

1

2m

0m

Find gradient of q 2m Find equation of q

2 2 y x

Solve p and q simultaneously for intersection

(0, 2)

Exam Type Exam Type QuestionsQuestions p

q

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Higher Outcome 1

Triangle ABC has vertices A(2, 2), B(12, 2) and C(8, 6).

a) Write down the equation of l1, the perpendicular bisector of

AB

b) Find the equation of l2, the perpendicular bisector of AC.

c) Find the point of intersection of lines l1 and l2.

7, 2Mid-point AB

Find mid-point AC

(5, 4) Find gradient of AC2

3m

Equation of perp. bisector AC

Gradient AC perp.3

2m 2 3 23y x

Point of intersection (7, 1)

7x Perpendicular bisector AB

Exam Type Exam Type QuestionsQuestions l1

l2

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Higher Outcome 1

A triangle ABC has vertices A(–4, 1), B(12,3) and C(7, –7).

a) Find the equation of the median CM.

b) Find the equation of the altitude AD.

c) Find the co-ordinates of the point of intersection of CM and AD

4, 2Mid-point AB

Equation of median CM

Gradient of perpendicular AD

Gradient BC 2m 1

2m

Equation of AD

3m Gradient CM (median)

3 14y x

Solve simultaneously for point of intersection(6, -4)

2 2 0y x

Exam TypeExam TypeQuestionsQuestions

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Higher Outcome 1A triangle ABC has vertices A(–3, –3), B(–1, 1) and C(7,–3).

a) Show that the triangle ABC is right angled at B.

b) The medians AD and BE intersect at M.

i) Find the equations of AD and BE. ii) Find find the co-ordinates of M.

2m Gradient AB

Product of gradients

Gradient of median AD

Mid-point BC 3, 1 1

3m Equation AD

1

2m Gradient

BC1

2 12

Solve simultaneously for M, point of intersection

3 6 0y x

Hence AB is perpendicular to BC, so B = 90°

Gradient of median BE

Mid-point AC 2, 3 4

3m Equation

AD3 4 1 0y x

51,

3

Exam Type Exam Type QuestionsQuestions

M

Education

Straight line