WAEC 2011 · Paper 2 · Q2

Coordinate geometry & circles

(a)
Find the equation of the line which passes through the point $(3, -2)$ and is perpendicular to the line $3x + 2y - 4 = 0$ .

Worked solution (try it first)

Rearrange: $2y = -3x + 4$ , so $y = -\frac32x + 2$ , with gradient $-\frac32$ .
The perpendicular gradient is $-1 \div \left(-\frac32\right) = \frac23$ .
$y + 2 = \frac23(x - 3)$ .
Multiply by 3: $3y + 6 = 2x - 6$ .
So $2x - 3y - 12 = 0$ .

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