Find the value of k, if the points A (7, -2), B (5, 1) and C (3, 2k) are collinear.

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Solution

If 3 points are collinear, then the area of the triangle formed by them is zero.

Area of triangle = $\frac{1}{2} | x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2}) | = 0$

$\Rightarrow x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2}) = 0$

$\Rightarrow 7 (1 - 2 k) + 5 (2 k + 2) + 3 (- 2 - 1) = 0$

$\Rightarrow 7 - 14 k + 10 k + 10 - 9 = 0$

$∴ k = 2$

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Similar questions

K

\begin{matrix} A (7, - 2) \end{matrix} \begin{matrix} B (5, 1) \end{matrix} \begin{matrix} C (3, K) \end{matrix}

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(\frac{2}{5}, \frac{1}{3}), (\frac{1}{2}, k)

(\frac{4}{5}, 0)

(\frac{2}{5}, \frac{1}{3})

(\frac{1}{2}, k)

(\frac{4}{5}, 0)

(\frac{1}{2}, 3)

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