If the points $(2, 3), (3, 5)$ and $(5, k)$ are collinear, then the value of $k$ is

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Solution

If the points are collinear then the area formed by the three points is $0$ , so
$\frac{1}{2} ∣ ∣ ∣ \begin{matrix} x_{1} & x_{2} & x_{3} & x_{1} y_{1} & y_{2} & y_{3} & y_{1} \end{matrix} ∣ ∣ ∣ = 0 \Rightarrow \frac{1}{2} ∣ ∣ ∣ \begin{matrix} 2 & 3 & 5 & 2 3 & 5 & k & 3 \end{matrix} ∣ ∣ ∣ = 0 \Rightarrow | (10 - 9) + (3 k - 25) + (15 - 2 k) | = 0 \Rightarrow | 1 + 3 k - 25 + 15 - 2 k | = 0 \Rightarrow k = 9$

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