CO-ORDINATE GEOMETRY

DISTANCE BETWEEN TWO POINTS- Found by using Pythagoras

In general points given are in the form (x1, y1), (x2, y2)

(x1, y1)

(x2, y2)

d

x2 – x1

y2 – y1

d2

= (x2 – x1)2

+ (y2 – y1)2

d2

= (x2 – x1)2

+ (y2 – y1)2

d = (x2 – x1)2

+ (y2 – y1)2

d = (x2 – x1)2

+ (y2 – y1)2

e.g. Calculate the distance between the points A = (-1, -3) and B = (2, 5)(x1, y1) (x2, y2)

d2

= (2 – -1)2

+ (5 – -3)2

d = (2 – -1)2

+ (5 – -3)2

d = (3)2

+ (8)2

d = 73d = 8·54 (2 d. p. )

Either co-ordinate can be labelled (x1, y1), but make sure you keep the correct x and y co-ordinates together!

Remember your integer laws!

MIDPOINT OF A LINE SEGMENT- The midpoint is the point exactly HALFWAY between the two end points

In general if the two points given are (x1, y1), (x2, y2), the co-ordinates of the midpoint are:

x1 + x2

2

'

y1 + y2

2

x1 + x2

2

'

y1 + y2

2

e.g. Find the midpoint of the line segment AB where A = (-1, -3) and B = (2, 5)(x1, y1) (x2, y2)

Midpoint = -1 + 2

2 ,

-3 + 5

2

= 12

, 1

Either co-ordinate can be labelled (x1, y1), but make sure you keep the correct x and y co-ordinates together!

GRADIENT OF A LINE- The gradient tells us how STEEP a line is

In general the gradient of the line joining the points (x1, y1), (x2, y2), is given by:

m =y2 – y1

x2 – x1

m =y2 – y1

x2 – x1

e.g. Find the gradient of the line joining the points A = (-1, -3) and B = (1, 5)(x1, y1) (x2, y2)

m = 5 – -31 – -1

m = 82

m = 4

Remember your integer laws!

Either co-ordinate can be labelled (x1, y1), but make sure you keep the y’s on top, the x‘s on the bottom and in the correct order!

Remember a horizontal line has a gradient of 0 and a vertical line’s gradient is undefined

EQUATION OF A LINE1. Any relation in the form y = mx + c gives a straight line where m = gradient

and c = y-intercept

e.g. Calculate the gradient and the y-intercept of the line

y =3

4 x +

5

4

y =3

4 x +

5

4

m = 34

c = 54

2. All straight lines can be written in the form ax + by + c = 0

e.g. Express the equation in the form ax + by + c = 0

y =1

2 x + 8

y =1

2 x + 8

2y = x + 16 Multiply ALL terms by denominator!

– x + 2y = 16– x + 2y – 16 = 0 Note: Other equivalent forms of equation are acceptable.

e.g. Find the gradient and y-intercept of the line

– 4x – 7y + 13 = 0

– 4x – 7y + 13 = 0 -7y + 13 = 4x-7y = 4x – 13

y = – 4x7

+ 137

Rearrange equation into the form y = mx + c

m = – 47

c = 137

FINDING AN EQUATION OF A LINE1. Given the gradient and a co-ordinate

- Use the point/gradient equation:

y – y1 = m(x – x1)

y – y1 = m(x – x1) e.g. Find the equation of the line that passes through the point (2, 3) with a gradient of 3/4. Express your answer in the form ax + by + c = 0

(x1, y1)

m

y – 3 = 34

(x – 2) Multiply ALL terms on the left by denominator!

4y – 12 = 3(x – 2)

4y – 12 = 3x – 6

-3x + 4y – 12 = -6

-3x + 4y – 6 = 0

Generally most equivalent forms of the equation will be acceptable as an answer and other methods of rearranging the equation are possible

Expand brackets and collect all terms on the left

or 4y – 3x – 6 = 0

2. Given two points (x1, y1), (x2, y2) on a line

- First calculate the gradient, then use the point/gradient equation:

e.g. Find the equation of the line joining the points (-4, 1) and (-3, -5). Express your answer in the form ax + by + c = 0

(x1, y1) (x2, y2)

m = -5 – 1-3 – -4

m = -61

m = -6

y – 1 = -6(x – -4)

It is up to you which point you use, but easiest to use positive numbers

y – 1 = -6(x + 4)

y – 1 = -6x – 24

6x + y – 1 = -24

6x + y + 23 = 0

m =y2 – y1

x2 – x1

m =y2 – y1

x2 – x1

y – y1 = m(x – x1)

y – y1 = m(x – x1)

PARALLEL LINES

- Have the same direction- Never meet- Have the same gradients so:

m1 = m2

m1 = m2

e.g. Find the equation of the line which passes through the point (2, -9) and is parallel to the line

5x + 2y – 3 = 0

5x + 2y – 3 = 0 First rearrange the equation to find the gradient.

2y = -5x + 3

y = – 5x2

+ 32

m = – 52

As the lines are parallel m1 = m2, and so we use the gradient found in the point/gradient equation

y – y1 = m(x – x1)

y – y1 = m(x – x1)

y – -9 = – 52

(x – 2)

2y + 18 = -5(x – 2)2y + 18 = -5x + 10

5x + 2y + 18 = 105x + 2y + 8 = 0

(x1, y1)

PERPENDICULAR LINES

- Meet at right angles- Gradients multiply to give -1 so

m1 m2 = – 1 and m1 =– 1

m2

or m2 =– 1

m1

m1 m2 = – 1 and m1 =– 1

m2

or m2 =– 1

m1

Note: Horizontal and vertical lines meet are perpendicular, but m1 × m2 has no value

e.g. Find the gradient of the line perpendicular to

2x + 3y – 4 = 0

2x + 3y – 4 = 0

3y = -2x + 4

y = – 2x3

+ 43

m1 = – 23

To find perpendicular gradient simply flip fraction over and change the sign!

m2 =-1

– 23

m2 = 32

e.g. Find the equation of the line that passes through (-1, 1) and is perpendicular to the line through the points (-3, 5) and (1, 4)

First find gradient

m = 4 – 51 – -3

m = -14

m =y2 – y1

x2 – x1

m =y2 – y1

x2 – x1

Find perpendicular gradient

(x1, y1) (x2, y2)

m1 = – 14

m2 = 4

m1 m2 = -1

Find equation

(x1, y1)

y – 1 = 4(x – -1)

y – 1 = 4(x + 1)

y – 1 = 4x + 4

-4x + y – 1 = 4

-4x + y – 5 = 0

y – y1 = m(x – x1)

y – y1 = m(x – x1)

or y – 4x – 5 = 0

e.g. Find the equation of the perpendicular bisector of the line joining the points (4, 1) and (0, 5)

First find gradient

(x1, y1) (x2, y2)

m = 5 – 10 – 4

m = 4-4

m = -1

Find perpendicular gradient

m1 = -1

m2 = 1

Find mid-point (bisector)

Midpoint = -4 + 0

2 ,

1 + 5

2

=(-2 , 3)

x1 + x2

2

'

y1 + y2

2

x1 + x2

2

'

y1 + y2

2

m =y2 – y1

x2 – x1

m =y2 – y1

x2 – x1

m1 m2 = -1

Find equation

y – 3 = 1(x – -2)

y – 3 = 1(x + 2)

y – 3 = x + 2

– x + y – 3 = 2

– x + y – 5 = 0

y – y1 = m(x – x1)

y – y1 = m(x – x1)

(x1, y1)

or y – x – 5 = 0

COLLINEAR POINTS- Points that lie on the same line- Gradients between any two pairs of points should be the same

e.g. Show that the points A = (-3, 4), B = (0, 2) and C = (6, -2) are collinear

Find gradient mAB

m = 2 – 40 – -3

m = -23

m =y2 – y1

x2 – x1

m =y2 – y1

x2 – x1

(x1, y1) (x2, y2)

Find gradient mAC

(x1, y1) (x2, y2)

m = -2 – 46 – -3

m = -69

m =y2 – y1

x2 – x1

m =y2 – y1

x2 – x1

As both gradients are equal, all three points must be collinear.

m = -23

POINTS OF INTERSECTION- Point where two lines cross- Found by using simultaneous equations

e.g. Find the point of intersection of the following pairs of straight lines

a) y = 2x – 5 y = -3x + 10

2x – 5 = -3x + 10

5x – 5 = 10+5+5

5x = 15÷5÷5

x = 3

Now we substitute x-value into first equation to find ‘y’ y = 1

+3x+3x

y = 2×3 – 5

Intersection point = (3, 1)

b) 3x – 5y = -7 -5x + 3y = 9 + ( )

-16y = -8÷-16÷-16

y = 0.5

Now we substitute y-value into either equation to find ‘x’

3x – 5 × 0.5 = -73x – 2.5 = -7

+2.5+2.53x = -4.5

Multiply the 1st equation by ‘5’ and the 2nd by ‘3’ then addto eliminate the x

× 5× 3

15x – 25y = -35 -15x + 9y = 27

÷3÷3x = 1.5

Intersection point = (1.5, 0.5)

APPLICATIONS- Generally require multiple steps of basic co-ordinate geometry techniques- It may be useful to draw a picture to help decide what steps to take

e.g. The points P(-5, -1), Q(-3, 5) and R(1, -3) are the vertices of a triangle. Prove that PQR is a right angled isosceles triangle.

x-5 -4 -3 -2 -1 1 2 3 4 5

y

-5-4-3-2-1

12345

1. Prove triangle is isosceles

- Need to prove 2 sides are of equal length.

PQ = (-3 – -5)2

+ (5 – -1)2

PQ = 40

PR = (1 – -5)2

+ (-3 – -1)2

PR = 40

QR = (1 – -3)2

+ (-3 – 5)2

QR = 80

- As two lengths are equal, triangle is isosceles.

2. Prove triangle is right angled.

- Need to show m1m2 = -1

mPQ = 5 – -1-3 – -5

mPQ = 62

mPR = -3 – -11 – -5

mPR = -26

mPQ mPR = 62

– 26

- As m1m2 = -1, lines are perpendicular and triangle is right angled.

P

Q

R