Find the values of $k,$ if the points $A (k + 1, 2 k), B (3 k, 2 k + 3)$ and $C (5 k + 1, 5 k)$ are collinear.

1

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\frac{1}{2}

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2

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2.5

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Solution

The correct options are
B $\frac{1}{2}$
C $2$
Let us assume that the points $A (x_{1}, y_{1}) = (k + 1, 2 k), B (x_{2}, y_{2}) = (3 k, 2 k + 3) & C (x_{3}, y_{3}) = (5 k + 1, 5 k)$ form a triangle.

Now, $a r e a (Δ A B C)$ with vertices $A (x_{1}, y_{1}), B (x_{2}, y_{2})$ and $C (x_{3}, y_{3})$ is given by

$A r e a (Δ A B C) = \frac{1}{2} [x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})]$

$∴ A r e a (Δ A B C) = \frac{1}{2} [(k + 1) (2 k + 3 - 5 k) + 3 k (5 k - 2 k) + (5 k + 1) (2 k - 2 k - 3)]$
$∴ A r e a (Δ A B C) = 2 k^{2} - 5 k + 2$

Since, $A, B, C$ are collinear $⟹ A r e a (Δ A B C) = 0$

$⟹ 2 k^{2} - 5 k + 2 = 0$

$⟹ (k - 2) (2 k - 1) = 0$

$⟹ k = 2, \frac{1}{2}$

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