For what values of k are the points $A (8, 1)$ , $B (3, - 2 k)$ and $C (k, - 5)$ collinear.

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Solution

Let $A (x_{1} = 8, y_{1} = 1), B (x_{2} = 3, y_{2} = - 2 k)$ & $C (x_{3} = k, y_{3} = - 5)$
be the given point
The given points are collinear if
$x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2}) = 0$
on puting values we get
$8 (- 2 k + 5) + 3 (- 5 - 1) + k (1 + 2 k) = 0$
$- 16 k + 40 - 18 + k + 2 k^{2} = 0$
$2 x^{2} - 15 x + 22 = 0$
$2 k^{2} - 4 k - 11 k + 22 = 0$
$2 k (k - 2) - 11 (k - 2) = 0$
$(2 k - 11) (k - 2) = 0$
$k = \frac{11}{2}, 2$
Hence for $k = \frac{11}{2}, 2$ given points are collinear.

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