Cambridge International AS and A Level Mathematics Pure Mathematics 1

Finding the equation of a line

P1^

2

EXAMPLE 2.6 Find the equation of the line with gradient 3 which passes through the point (2, −4).

SOLUTION

Using y − y 1 = m(x − x 1 )

⇒ y − (−4) = 3(x − 2)

⇒ y + 4 = 3 x − 6

⇒ y = 3 x− 10.

(ii) Given two points, (x 1 , y 1 ) and (x 2 , y 2 )

The two points are used to find the

gradient:

m

yy

=xx

21

21

–

–.

This value of m is then substituted in

the equation

y − y 1 = m (x − x 1 ).

This gives

yy

yy

• xx xx

–

(^1) – –.
21
21 1

= ()

Rearranging the equation gives

yy

yy

xx

xx

yy

xx

yy

xx

–

–

–

–

–

–

–

–

1

21

1

21

1

1

21

21

= or =

EXAMPLE 2.7 Find the equation of the line joining (2, 4) to (5, 3).

SOLUTION

Taking (x 1 , y 1 ) to be (2, 4) and (x 2 , y 2 ) to be (5, 3), and substituting the values in

yy

yy

xx

xx

–

–

–

–

1

21

1

21

=

gives

yx–

• 
  

–

–.

4

34

2

= 52

This can be simplified to x + 3 y − 14 = 0.

●?^ Show that the equation of the line in figure 2.19

can be written

x

a

y

b

+=1.

y

x

(x 2 , y 2 )

(x 1 , y 1 ) (x, y)

O

Figure 2.18

(a, 0)

O

y

x

(0, b)

Figure 2.19