Question

In the following, find the value of ‘a’ for which the given points are collinear (2, 3), (4, a) and (6, – 3). (2 marks)

Answer 1

$LetthepointsbeA(2,3),B(4,a)andC(6,–3).Sincethegivenpointsarecollinear.Areaofatriangle=0(0.5mark)Areaof\Delta ABC=\frac{1}{2}\left[\right(x_1{y}_2+x_2{y}_3+x_3{y}_1)-(x_2{y}_1+x_3{y}_2+x_1{y}_3\left)\right](0.5mark)\frac{1}{2}\left[\right(2a-12+18)-(12+6a-6\left)\right]=02a+6–(6+6a)=02a+6–6–6a=0–4a=0\Rightarrow a=0\therefore Thevalueofa=0\left(1mark\right)$