JAMB 2016 · UTME · Q39

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

Worked solution (try it first)
  1. Gradient of the first line: −2−46−p=−66−p\dfrac{-2 - 4}{6 - p} = \dfrac{-6}{6 - p}.
  2. Gradient of the second line: 3−p−1−2=p−33\dfrac{3 - p}{-1 - 2} = \dfrac{p - 3}{3}.
  3. Perpendicular gradients multiply to −1-1: −6(p−3)3(6−p)=−1\dfrac{-6(p - 3)}{3(6 - p)} = -1, so 2(p−3)=6−p2(p - 3) = 6 - p.
  4. Then 3p=123p = 12, so p=4p = 4, option C.
  5. Check: the gradients are −3-3 and 13\frac13, and −3×13=−1-3 \times \frac13 = -1.

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