Cambridge International Mathematics

306 Straight lines (Chapter 14)

This suggests that: ² for gradient

a

b

the general form of the line is ax¡by=d

² for gradient ¡

a

b

the general form of the line is ax+by=d.

The constant termdon the RHS is obtained by substituting the coordinates of any point which lies on the

line.

Example 6 Self Tutor

Find the equation of a line:

a with gradient^34 , that passes through(5,¡2)

b with gradient ¡^34 , that passes through(1,7).

a The equation is 3 x¡ 4 y= 3(5)¡4(¡2)

) 3 x¡ 4 y=23

b The equation is 3 x+4y= 3(1) + 4(7)

) 3 x+4y=31

EXERCISE 14D.2

1 Find the equation of a line:

a through(4,1)with gradient^12 b through(¡ 2 ,5)with gradient^23

c through(5,0)with gradient^34 d through(3,¡2)with gradient 3

e through(1,4)with gradient¡^13 f through(2,¡3)with gradient¡^34

g through(3,¡2)with gradient¡ 2 h through(0,4)with gradient¡ 3.

2 We can use the reverse process to question 1 to write down the gradient of a line given in general form.

Find the gradient of the line with equation:

a 2 x+3y=8 b 3 x¡ 7 y=11 c 6 x¡ 11 y=4

d 5 x+6y=¡ 1 e 3 x+6y=¡ 1 f 15 x¡ 5 y=17

3 Explain why:

a any line parallel to 3 x+5y=2 has the form 3 x+5y=d

b any line perpendicular to 3 x+5y=2 has the form 5 x¡ 3 y=d.

4 Find the equation of a line which is:

a parallel to the line 3 x+4y=6 and passes through(2,1)

b perpendicular to the line 5 x+2y=10 and passes through(¡ 1 ,¡1)

c perpendicular to the line x¡ 3 y+6=0 and passes through(¡ 4 ,0)

d parallel to the line x¡ 3 y=11 and passes through(0,0).

5 2 x¡ 3 y=6 and 6 x+ky=4 are two straight lines.

a Write down the gradient of each line. b Findkif the lines are parallel.

c Findkif the lines are perpendicular.

With practice you can

write down the equation

very quickly, but you

can still use the

method if

you need.

ymxc¡¡ ¡¡=+

IGCSE01

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Y:\HAESE\IGCSE01\IG01_14\306IGCSE01_14.CDR Friday, 26 September 2008 12:39:14 PM PETER