Finding the equation of a line
P1^
2
SOLUTION
The line 2y = x + 12 has gradient^12 and passes through the point (0, 6)
(since dividing by 2 gives y = 12 x + 6).
The line y = 4 x − 10 has gradient 4 and passes through the point (0, −10).
The other two sides are lines with gradients^12 and 4 which pass through (0, 0),
i.e. y = 12 x and y = 4 x.
EXAMPLE 2.10 Find the equation of the perpendicular bisector of the line joining P(−4, 5) to
Q(2, 3).
SOLUTION
y = 4x – 10
2 y = x + 12
(0, –10)
x
y
O
(0, 6)
The dashed lines
are the other
two sides.
Figure 2.21
y
x
P(–4, 5)
Q(2, 3)
R
O
Figure 2.22