JAMB 2000 · UME · Q28

$3y = 4x - 1$ and $Ky = x + 3$ are equations of two straight lines. If the two lines are perpendicular to each other, find $K$ .

Worked solution (try it first)

$3y = 4x - 1$ gives $y = \frac43x - \frac13$ , so its gradient is $\frac43$ .
$Ky = x + 3$ gives $y = \frac1Kx + \frac3K$ , so its gradient is $\frac1K$ .
Perpendicular gradients multiply to $-1$ : $\frac43 \times \frac1K = -1$ , so $\frac1K = -\frac34$ .
Turn it upside down: $K = -\frac43$ , option A.