1: Straight Lines and 1: Straight Lines and GradientsGradients

““Teach A Level Maths”Teach A Level Maths”

Vol. 1: AS Core Vol. 1: AS Core ModulesModules

4

2

22

4 my

Finding the Gradient

( 2, 3 )

( 0, 1)

3 ( 1) 4

2 0 2

02

)1(3m

2

4m 2m

12

12

xx

yym

12 xx

12 yy

)3,2( ),( 22 yx

),( 11 yx )1,0(

),( 22 yx

The gradient of the straight line joining the pointsand),( 11 yx ),( 22 yx

is

e.g. Find the gradient of the straight line joining the points and)1,0( )3,2(

22

4 mm

)1(

12

12

xx

yym

m 2 03

To use this formula, we don’t need a diagram!

),( 11 yxSolution:

12

12

xx

yym

Equation of a Straight Line

The gradient of the straight line joining the pointsand),( 11 yx ),( 22 yx

is12

12

xx

yym

Equation of a Straight Line

Find the gradient of the line joining the points A(3,–2) and B(–5,6)

12

12

xx

yym

x2 = –5 y2 = 6x1= 3 y1= –2

Equation of a Straight Line

Find the gradient of the line joining the points A(3,–2) and B(–5,6)

12

12

xx

yym

x2 = –5 y2 = 6

35

)2(6

m 35

26

8

8

1

x1 = 3 y1= –2

Equation of a Straight Line

c is the point where the line meets the y-axis, the y-intercept

and y-intercept, c = 2

1e.g. has gradient m = 12 xy

cmxy • The equation of a straight line ism is the gradient of the line

gradient = 2

x

12 xy

intercept on y-axis

Equation of a Straight Line

gradient = 2

x

12 xy

intercept on y-axis

( 4, 7 )x

• The coordinates of any point lying on the line satisfy the equation of the line

showing that the point ( 4,7 ) lies on the line.

71)4(2 yye.g. Substituting x = 4 in gives12 xy

Equation of a Straight Line Finding the equation of a straight line when

we know

e.g.Find the equation of the line with gradient passing through the point

)3,1( 2

• its gradient, m and • the coordinates of a point on the line

(x1,y1).

Solution:

y m and x3, 2 1

So, 52 xy

y m x c c3 2 1

y x2 5

Using , m is given, so we can find c bysubstituting for y, m and x.

y m x c

(-1, 3)

52 xy

x

c3 2 c3 2

Add 2 to both sidesC = 5

Equation of a Straight Line

Solution: First find the gradient

We could use the 2nd point,(-1, 3) instead of (2, -3)

Using the formula when we are given two points on the line

cmxy

e.g. Find the equation of the line through the points )3,1()3,2( and

12

12

xx

yym

12 xy

2)1(

)3(3

m3

6

m 2 m

Now use with

cmxy 3 and2 yx

c 223

1 c

),( 11 yx 2 2( , )x y

-3 = -4 + c

Add 4 to both sides

Equation of a Straight Line

SUMMARY

Equation of a straight line

Gradient of a straight line

12

12

xx

yym

y mx c

where and are points on the line

),( 11 yx ),( 22 yx

where m is the gradient and c is the intercept on the y-axis

Equation of a Straight Line

2. Find the equation of the line through the points )4,1()2,1( and

Exercise1. Find the equation of the line with gradient 2

which passes through the point . )1,4( Solution: 92 xySo,

Solution:

y ym

x x2 1

2 1

m

4 23

1 ( 1)

13 xySo,

Equation of a Straight Line

We sometimes rearrange the equation of a straight line so that zero is on the right-hand side ( r.h.s. )

We must take care with the equation in this form.

e.g. can be written as

12 xy 012 yx

e.g. Find the gradient of the line with equation

0734 yxSolution: Rearranging to the form :

cmxy

0734 yx 743 xy

3

7

3

4

x

y

)( cmxy

so the gradient is 3

4