Question

Find the value of k if $\left(\frac{2}{5},\frac{1}{3}\right)$ , $\left(\frac{1}{2},k\right)$ and $\left(\frac{4}{5},0\right)$ are collinear points .

Answer 1

Given: Points $\left(\frac{2}{5},\frac{1}{3}\right)$ , $\left(\frac{1}{2},k\right)$ and $\left(\frac{4}{5},0\right)$ are collinear.

So, slope of the line joining $\left(\frac{2}{5},\frac{1}{3}\right)$ and $\left(\frac{1}{2},k\right)$ = slope of the line joining $\left(\frac{1}{2},k\right)$ and $\left(\frac{4}{5},0\right)$

$$\begin{gathered} \mathrm{i}.\mathrm{e}.,\frac{y_2-y_1}{x_2-x_1}=\frac{y_3-y_2}{x_3-x_3} \\ \Rightarrow \frac{k-{\displaystyle \frac{1}{3}}}{{\displaystyle \frac{1}{2}}-{\displaystyle \frac{2}{5}}}=\frac{0-k}{{\displaystyle \frac{4}{5}}-{\displaystyle \frac{1}{2}}} \\ \Rightarrow \frac{k-{\displaystyle \frac{1}{3}}}{{\displaystyle \frac{1}{10}}}=\frac{-k}{{\displaystyle \frac{3}{10}}} \\ \Rightarrow \frac{3}{10}\left(k-{\displaystyle \frac{1}{3}}\right)=\frac{1}{10}\left(-k\right) \\ \Rightarrow 3\left(k-{\displaystyle \frac{1}{3}}\right)=-k \\ \Rightarrow 3k-1=-k \\ \Rightarrow 3k+k=1 \\ \Rightarrow 4k=1 \\ \Rightarrow k=\frac{1}{4} \end{gathered}$$