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WAEC 2022 · Paper 2 · Q3

Three points P(−3,2)P(-3, 2), Q(7,k)Q(7, k) and R(2,−5)R(2, -5) are in a plane. The line through the points PP and QQ has the same gradient as 3x+5y−2=03x + 5y - 2 = 0. Find the:

  1. (a)

    value of kk;

  2. (b)

    distance between QQ and RR (1 d.p.).

Try it on a graph

P, Q and R plotted. PQ is parallel to 3x + 5y − 2 = 0 (grey).

Worked solution (try it first)

(a)

  1. 3x+5y−2=03x + 5y - 2 = 0 gives 5y=−3x+25y = -3x + 2, so y=−35x+25y = -\frac35x + \frac25 and the gradient is −35-\frac35.
  2. Gradient of PQ=k−27−(−3)PQ = \frac{k - 2}{7 - (-3)}
    =k−210= \frac{k - 2}{10}.
  3. The gradients are equal: k−210=−35\frac{k - 2}{10} = -\frac35, so k−2=−6k - 2 = -6 and k=−4k = -4.

(b)

  1. With Q(7,−4)Q(7, -4) and R(2,−5)R(2, -5): ∣QR∣=(7−2)2+(−4−(−5))2|QR| = \sqrt{(7 - 2)^2 + (-4 - (-5))^2}
    =25+1= \sqrt{25 + 1}
    =26= \sqrt{26}
    ≈5.1\approx 5.1 units.

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