Question

In each of the following find the value of $k$, for which the points are collinear. $\left(8,1\right),\left(k,-4\right)$ and $\left(2,-5\right)$

Answer 1

Find the value $k$

For collinear point, area of triangle formed by them is always zero

Let point $\left(8,1\right),\left(k,-4\right)$ are $\left(2,-5\right)$ are vertices of triangle.

$$\begin{gathered} \mathrm{Area}\mathrm{of}\mathrm{triangle}=\frac{1}{2}\left|\begin{array}{ccc}{\mathrm{a}}_1& {\mathrm{b}}_1& 1\\ {\mathrm{a}}_2& {\mathrm{b}}_2& 1\\ {\mathrm{a}}_3& {\mathrm{b}}_3& 1\end{array}\right| \\ \Rightarrow \mathrm{Area}\mathrm{of}\mathrm{triangle}=\frac{1}{2}\left|\begin{array}{ccc}8& 1& 1\\ k& -4& 1\\ 2& -5& 1\end{array}\right| \\ \Rightarrow \mathrm{Area}\mathrm{of}\mathrm{triangle}=\frac{1}{2}\left[8\left(-4+5\right)+1\left(2-k\right)+1\left(-5\mathrm{k}+8\right)\right] \\ \Rightarrow 0=8+2-\mathrm{k}-5\mathrm{k}+8\left[\because \mathrm{For}\mathrm{collinear}\mathrm{point}\mathrm{area}\mathrm{of}\mathrm{triangle}\mathrm{is}0\right] \\ \Rightarrow 0=18-6\mathrm{k} \\ \Rightarrow \mathrm{k}=3 \end{gathered}$$

Hence value of $k$ is $3$