The Straight Line

y

x

y

x

All straight lines have an equation of the form y mx c

m = gradient y axis intercept

C C

+ ve gradient

- ve gradient

1 1( , )A x y

2 2( , )B x y

Vertical HeightGradient =

Horizontal Distancem

2 1y y

2 1x x

2 1

2 1

y y

x x

Undefined and zero gradient

Gradient is a measure of slope. If a line has zero gradient it has zero slope.

A line with zero slope is horizontal.

y

x

2

2

4

4

6

6

8

8

10

10

– 2

– 2

– 4

– 4

– 6

– 6

– 8

– 8

– 10

– 10

2

2

4

4

6

6

8

8

10

10

– 2

– 2

– 4

– 4

– 6

– 6

– 8

– 8

– 10

– 10

Consider two points on this graph.

( 4,4) (6,4)2 1

2 1

y ym

x x

4 4

6 4

0

The equation of the line is 4y

All horizontal lines have an equation of the form y c

y

x

2

2

4

4

6

6

8

8

10

10

– 2

– 2

– 4

– 4

– 6

– 6

– 8

– 8

– 10

– 10

2

2

4

4

6

6

8

8

10

10

– 2

– 2

– 4

– 4

– 6

– 6

– 8

– 8

– 10

– 10

(4, 6)

(4,8)

Consider two points on this graph.

2 1

2 1

y ym

x x

8 6

4 4

(undefined)

The equation of the line is 4x

All vertical lines have an equation of the form x a

x

y

yx

From the diagram we can see that

tany

x

m

tanm

Note that is the angle the line makes with the positive direction of the x axis.

Collinearity

Two lines can either be:

A

B C

At an angle Parallel and Distinct

A

B

C

D

A

B

C

Parallel and form a straight line

Points that lie on the same straight line are said to be collinear.

To prove points are collinear:

1. Show that two pairs of points have the same gradient. (parallel)

2. If the pairs of points have a point in common they MUST be collinear.

1. Prove that the points P(-6 , -5), Q(0 , -3) and R(12 , 1) are collinear.

3 5

0 6pqm

2

6

1

3

1 3

12 0qrm

4

12

1

3

As the gradients of PQ and QR are equal PQ is parallel to QR.

Since Q is a point in common to PQ and QR, the points P, Q and R

are collinear.

Page 3 Exercise 1B

Perpendicular Lines

x

y

( , )a b

If we rotate the line through 900.

Perpendicular Lines

x

y

( , )a b

If we rotate the line through 900.

Perpendicular Lines

x

y

( , )a b

If we rotate the line through 900.

Perpendicular Lines

x

y

( , )a b

B( , )b a

O

A

This is true for all perpendicular lines.

If two lines with gradients m1 and m2 are perpendicular then

0

0OA

bm

a

b

a

0

0OB

am

b

a

b

1OA OB

b am m

a b

1 2 1m m Conversely, if then the lines with gradients m1 and m2

are perpendicular.

1 2 1m m

1. If P is the point (2,-3) and Q is the point (-1,6), find the gradient of the line perpendicular to PQ.

6 3

1 2PQm

9

3 3

To find the gradient of the line perpendicular to PQ we require thenegative reciprocal of –3.

Remember: 1a b

b a

3Since 3

1 1

The negative reciprocal would be 3

1The gradient of the line perpendicular to PQ is .

3

2. Triangle RST has coordinates R(1,2), S(3,7) and T(6,0). Show that the triangle is right angled at R.

S

T

R

7 2

3 1RSm

5

2

0 2

6 1RTm

2

5

5 21

2 5RS RTm m

Since 1, RS is perpendicular to RT.RS RTm m

Hence the triangle is right angled at R.

Equation of a Straight Line

All straight lines have an equation of the form y mx c

x

y

( , )P x y

A(0,C)

m

P(x,y) is any point on the line except A.

For every position P the gradient of AP is

0

y cm

x

y c

x

y c mx

y mx c

1. What is the equation of the line with gradient 2 passing through the point (0,-5)?

2 and 5m c

the equation of the line is 2 5y x

2. Find the gradient and the y intercept of the line with equation 4 3 2x y

Rearranging gives: 3 4 2y x 4 2

3 3y x

4 2Gradient is - and the axis intercept is .

3 3y

3. Show that the point (2,7) lies on the line 4 1.y x

When 2,x 4 2 1y 7

Because (2,7) satisfies the equation y = 4x – 1, the point must lie on the line.

General Equation of a Straight Line

0 is used as an alternative to Ax By C y mx c

0 is the GENERAL EQUATION of a straight line.Ax By C

1. Rearrange 2 5 into the form 0

and identify the values of A, B and C.

y x Ax By C

2 5 0x y 2, 1, 5A B C

42. Rearrange into the form 0

3and identify the values of A, B and C.

xy Ax By C

3 4y x 4, 3, 0A B C 4 3 0x y

3. Rearrange 7 into the form 0

and identify the values of A, B and C.

x Ax By C

7 0x 1, 0, 7A B C

Finding the equation of a Straight Line

To find the equation of a straight we need•A Gradient•A Point on the line

The equation of a straight line with gradient m passing through (a,b) is

( )y b m x a

x

y

( , )P x y

A(a,b)

m

P(x,y) is any point on the line except A.

For every position P the gradient of AP

y bm

x a

1

m y b

x a

( )y b m x a

11. Find the equation of the straight line passing through (5, 2) with gradient .

2

1(5, 2),

2P m

1( 2) ( 5)

2y x

1 52

2 2y x

1 9

2 2y x or 2 9y x

Equating with ( )y b m x a

2. Find the equation of the line passing through P(-2,0) and Q(1,6).

6 0

1 2PQm

2

Using point P

0 2( 2)y x

2 4y x

But what if we used point Q?

6 0

1 2PQm

2

Using point Q

6 2( 1)y x

2 4y x

6 2 2y x

Regardless of the point you use the equation of the straight line will ALWAYS be the same as both points lie on the line.

Lines in a Triangle

1. The Perpendicular Bisector.

A perpendicular bisector will bisect a line at 900 at the mid point.

The point of intersection is called the Circumcentre.

1. A is the point (1,3) and B is the point (5,-7). Find the equation of the perpendicular bisector of AB.

To find the equation of any straight line we need a point and a gradient.

Mid-point of AB = 1 5 3 ( 7)

,2 2

3, 2

7 3

5 1ABm

10 5

4 2

2

5m 1 2since 1m m

22 3

5y x

2 16

5 5y x

2. The Altitude.

An altitude of a triangle is a line from a vertex perpendicular to the opposite side. A triangle has 3 altitudes.

The point of intersection is called the Orthocentre.

3. The Median of a Triangle.

The median of a triangle is a line from a vertex to the mid point of the opposite side. A triangle has 3 medians.

The point of intersection is called the centroid

A further point of information regarding the centroid.

2

1

The centroid is a point of TRISECTION of the medians. It divides each median in the ratio 2:1.

1. F, G and H are the points (1,0), (-4,3) and (0,-1) respectively. FJ is a median of triangle FGH and HR is an altitude. Find the coordinates of the point of intersection D, of FJ and HR.

(Draw a sketch – It HELPS!!)

F

G

H

MEDIAN

J

4 0 3 1,

2 2J

2,1

1 0

2 1FJm

1

3

10 1

3y x

1 1

3 3y x

ALTITUDE

F

G

H

JR

H(0,-1)

3 0

4 1FGm

3

5

5

3HRm

51

3y x

51

3y x

The point D occurs when;

D

1 1 51

3 3 3x x

42

3x

2

3x

5 21

3 3y

1

9

2 1,

3 9D