1 Straight Lines and Gradients

1: Straight Lines and 1: Straight Lines and GradientsGradients

© Christine Crisp

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

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

Straight Lines and Gradients

Module C1

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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


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


Notice that to find c, the equation has been solved from right to left. This takes a bit of practice but reduces the chance of errors.

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.

Solution: 12,3 xmy and

So, 52 xy

ccmxy )1(23c 5

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

cmxy

(-1, 3)

52 xy

x


If we don’t know the gradient, we have to find it using two points on the line.

We develop the formula by reminding ourselves about the meaning of a gradient.

To do this, we can use a formula.


4

2

e.g.

22

4 mm


( 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(


12

12

xx

yym

m

),( 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(2 0

3

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

),( 11 yxSolution:

12

12

xx

yym


To find the equation of a straight line given 2 points on the line.

Solution: First find the gradient:

e.g. Find the equation of the line through the points

)3,1()3,2( and

12

12

xx

yym

Now

cmxy on the line:

)3,2( c )2(23c 1

Equation of line is

12 xy

3

6

m

2)1(

)3(3

m

2 m

cxy 2


SUMMARY

Equation of a straight line

Gradient of a straight line

12

12

xx

yym

cmxy

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


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:

cxy 2c )4(21)1,4( line on c 9

92 xySo,

Solution:

12

12

xx

yym

3)1(1

24

m

cmxy

cmxy cxy 3c )1(32)2,1( line on c 1

13 xySo,


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


Parallel and Perpendicular Lines

They are parallel if 12 mm

They are perpendicular if 1

21

mm

If 2 lines have gradients and , then:1m 2m


e.g. 1 Find the equation of the line parallel towhich passes through the point )3,1(

12 xy

Solution: The given line has gradient 2. Let 21 mFor parallel lines,

22 m12 mm

is the equation of any line parallel to

Using cmxy cxy 2

12 xyon the line

)3,1( c )1(23

c 1

12 xy


We don’t usually leave fractions ( or decimals ) in equations. So, multiplying by 2:

e.g.Find the equation of the line perpendicular to passing through the point . 12 xy )4,1(

Solution: The given line has gradient 2. Let

21 m

Perpendicular lines:2

12 m

12

1

mm

Equation of a straight line: cmxy on the line

)4,1( c )1(2

14 c

2

9

2

9

2

1 xy

92 xy 092 yxor


• If the gradient isn’t given, find the gradient using

Method of finding the equation of a straight line:

• Substitute for y, m and x in intoto find c.

cmxy

either parallel lines: 12 mm

or 2 points on the line:

12

12

xx

yym

or perpendicular lines:1

21

mm

SUMMARY

Straight Lines and GradientsExercise

Solution:

12 xy

c )2(23)3,2( line on c 112 xySo,

Solution: 2

321032 xyxy

21

1 m

012xy

12

1

mm cxy 2

c 22)2,1( line on c 0xy 2So,

1. Find the equation of the line parallel to the line

which passes through the point .

)3,2( 012 xy2 mParallel line

iscxy 2

2. Find the equation of the line through the point (1, 2), perpendicular to the line 032 xy

22 m So,


A Second Formula for a Straight Line ( optional )Let ( x, y ) be any point on the line

1xx

1yy

1

1

xx

yym )( 11 xxmyy

Let be a fixed point on the line

),( 11 yx

),( yxx

),( 11 yx x


Solution: First find the gradient

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

To use the formula we need to be given

either: one point on the line and the gradient

or: two points on the line

)( 11 xxmyy

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 )( 11 xxmyy 32 11 yx and

)2)(2()3( xy423 xy


The following slides contain repeats of information on earlier slides, shown without colour, so that they can be printed and photocopied.For most purposes the slides can be printed as “Handouts” with up to 6 slides per sheet.


They are parallel if 12 mm

They are perpendicular if 1

2

1

mm

If 2 lines have gradients and , then:1m 2m

Equation of a straight line

Gradient of a straight line

12

12

xx

yym

cmxy

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

SUMMARY


Solution: First find the gradient:

e.g. Find the equation of the line through the points

)3,1()3,2( and

12

12

xx

yym

2)1(

)3(3

m

3

6

m 2 m

Now

cmxy cxy )(2on the line:

)3,2( c )2(23c 1

Equation of line is

12 xy


We don’t usually leave fractions ( or decimals ) in equations. So, multiplying by 2:

e.g.Find the equation of the line perpendicular to passing through the point . 12 xy )4,1(

Solution: The given line has gradient 2. Let

21 m

Perpendicular lines:2

12 m

12

1

mm

Equation of a straight line: cmxy on the line

)4,1( c )1(2

14 c

2

9

2

9

2

1 xy

92 xy 092 yxor

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1 Straight Lines and Gradients