Q22 of 157 Page 753

If the points P(–3, 9), Q(a, b) and R(4, –5) are collinear and a + b = 1, find the values of a and b.


To show that the points are collinear, we show that the area of triangle is equilateral = 0


Δ = 0


Δ = 1/2{x1(y2−y3) + x2(y3−y1) + x3(y1−y2)}


⇒ Δ = 1/2{–3 (b + 5) + a (–5–9) + 4 (9–b)} = 0


⇒ –3b–150–14a + 36–4b = 0


2a + b = 3


Now solving a + b =1 and 2a + b = 3 we get a = 2 and b = −1.


Hence a = 2, b = –1


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