Question 2 (ii)
Find the value of 'k', for which the following points are collinear.
(ii) (8, 1), (k, -4), (2, -5)
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Solution
For collinear points, area of triangle formed by them is zero.
Therefore, for points (8, 1), (k, - 4) and (2, - 5), area = 0
Area of a triangle
= 12{x1(y2−y3)+x2(y3−y1)+x3(y1−y2)}
=12[8{−4−(−5)}+k{(−5)−(1)}+2{1−(−4)}]=0
=8−6k+10=0
=6k=18
=k=3