JAMB 2014 · UTME · Q36

Find the equation of the straight line through $(-2, 3)$ and perpendicular to $4x + 3y - 5 = 0$ .

Worked solution (try it first)

Rearrange $4x + 3y - 5 = 0$ : $y = -\frac43x + \frac53$ , so its gradient is $-\frac43$ .
The perpendicular gradient is the negative reciprocal, $\frac34$ .
Through $(-2, 3)$ : $y - 3 = \frac34(x + 2)$ .
Multiply by 4: $4y - 12 = 3x + 6$ .
Collect terms: $3x - 4y + 18 = 0$ , option B.