For what value of 'k' are the following points (7,-2), (5, 1), (3, k) collinear ?

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Solution

The correct option is D
4

Let $A (7, - 2), B (5, 1) & C (3, k)$ be the vertices of a triangle.

Condition for collinear points, $a r (△ A B C) = 0$
Area of triangle having coordinates $(x_{1}, y_{1}), (x_{2}, y_{2}), (x_{3}, y_{3})$ is
$\frac{1}{2} \times | x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2}) |$

$∴ \frac{1}{2} | 7 (1 - k) + 5 (k + 2) + 3 (- 2 - 1) | = 0 7 - 7 k + 5 k + 10 + (- 6) - 3 = 0 17 - 9 + 5 k - 7 k = 0 8 - 2 k = 0 \Rightarrow 2 k = 8 \Rightarrow k = \frac{8}{2} ∴ k = 4$

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