JAMB 2001 · UME · Q30

Find the value of $p$ if the line joining $(p, 4)$ and $(6, -2)$ is perpendicular to the line joining $(2, p)$ and $(-1, 3)$ .

Worked solution (try it first)

Gradient of the first line: $\dfrac{-2 - 4}{6 - p} = \dfrac{-6}{6 - p}$ .
Gradient of the second line: $\dfrac{3 - p}{-1 - 2} = \dfrac{p - 3}{3}$ .
Perpendicular gradients multiply to $-1$ : $\dfrac{-6(p - 3)}{3(6 - p)} = -1$ , so $2(p - 3) = 6 - p$ .
Then $3p = 12$ , so $p = 4$ , option C.