coordinate geometry © christine crisp. straight lines and gradients c is the point where the line...

Coordinate geometryCoordinate geometry

© Christine Crisp

Straight Lines and Gradients

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

Straight Lines and Gradients

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

Straight Lines and Gradients

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

Straight Lines and Gradients

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

Straight Lines and Gradients

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

Straight Lines and Gradients

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

Straight Lines and Gradients

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

Straight Lines and Gradients

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

Straight Lines and Gradients

• 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 Gradients

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

Straight Lines and Gradients

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

Finding the Mid-Point of AB

The mid-point is the average of the end points:

M X

)2,2( A

)3,1(B

2

1

2

12

Mx

2

1

2

32

My

Finding the Mid-Point of AB

A formula for the Mid-Point of AB

M X

or

)2,2( A ),( 11 yxA

)3,1(B ),( 22 yxB

,2

21 xxx M

221 yy

y M

The mid-point is the average of the end points:

2

,2

2121 yyxxM is

Finding the Mid-Point of AB

Solution:

Exercise

Find the mid-point, M of the line joining to

)3,4( )2,1(

2

,2

2121 yyxxM is

2

23

2

)1(4,isM

2

1,

2

3

Length of the line joining A to B

Using Pythagoras’ theorem:

)2,2( A

)3,1(B

3)2(1

5)2(3

3453 22 AB

Length of the line joining A to B

A formula for the length of the line joining A to B

Using Pythagoras’ theorem:

)2,2( A ),( 11 yxA

)3,1(B ),( 22 yxB

12 xx

12 yy

212

212 )()( yyxxAB

Straight Lines and Gradients

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.

Straight Lines and Gradients

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

Straight Lines and Gradients

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

Straight Lines and Gradients

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