Question

Find the value of k if (1, 2), $\left(\frac{1}{2},3\right)$ and (0,k) are collinear points.

Answer 1

Given: Points (1, 2), $\left(\frac{1}{2},3\right)$ and (0,k) are collinear.
So, slope of the line joining (1, 2) and $\left(\frac{1}{2},3\right)$ = slope of the line joining $\left(\frac{1}{2},3\right)$ and (0, k )

$$\begin{gathered} \mathrm{i}.\mathrm{e}.,\frac{y_2-y_1}{x_2-x_1}=\frac{y_3-y_2}{x_3-x_3} \\ \Rightarrow \frac{3-2}{{\displaystyle \frac{1}{2}}-1}=\frac{k-3}{0-{\displaystyle \frac{1}{2}}} \\ \Rightarrow \frac{1}{-{\displaystyle \frac{1}{2}}}=\frac{k-3}{-{\displaystyle \frac{1}{2}}} \\ \Rightarrow 1=k-3 \\ \Rightarrow k=1+3 \\ \Rightarrow k=4 \end{gathered}$$