Question

For what values of k are the points A(8, 1), B(3, −2k) and C(k, −5) collinear? [CBSE 2015]

Answer 1

Let A(x₁ = 8, y₁ = 1), B(x₂ = 3, y₂ = −2k) and C(x₃ = k, y₃ = −5) be the given points.
The given points are collinear if
$$\begin{gathered} x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)=0 \\ \Rightarrow 8\left(-2k+5\right)+3\left(-5-1\right)+k\left(1+2k\right)=0 \\ \Rightarrow -16k+40-18+k+2k^2=0 \\ \Rightarrow 2k^2-15k+22=0 \end{gathered}$$
$$\begin{gathered} \Rightarrow 2k^2-11k-4k+22=0 \\ \Rightarrow k\left(2k-11\right)-2\left(2k-11\right)=0 \\ \Rightarrow \left(k-2\right)\left(2k-11\right)=0 \\ \Rightarrow k=2\mathrm{or}k=\frac{11}{2} \end{gathered}$$
Hence, $k=2\mathrm{or}k=\frac{11}{2}$.