If the points $( \frac{2}{5},\ \frac{1}{3}),\ ( \frac{1}{2},\ k)$ and $( \frac{4}{5},\ 0)$ are collinear, then find the value of $k$.

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

To do: To find the value of $k$.

When three points A, B and C are collinear, slope of line

joining any two points (say AB) = slope of line joining any other two points (say BC).

Slope of the line passing through points $(x_1, y_1)$ and $(x_2,\ y_2) =\frac{y_2-y_1}{x_2-x_1}$ 

Slope of the line passing through $( \frac{2}{5},\ \frac{1}{3})$ and $( \frac{1}{2},\ k)=$ Slope of line passing through $( \frac{1}{2},\ k)$ and $( \frac{4}{5},\ 0)$

  

$\Rightarrow \frac{k-\frac{1}{3}}{\frac{1}{2}-\frac{2}{5}}=\frac{0-k}{\frac{4}{5}-\frac{1}{2}}$

$\Rightarrow \frac{\frac{3k-1}{3}}{\frac{1}{10}}=\frac{-k}{\frac{3}{10}}$

$\Rightarrow \frac{10( 3k-1)}{3}=-\frac{10k}{3}$

$\Rightarrow 30k-10=-10k$

$\Rightarrow 40k=10$

$\Rightarrow k=\frac{10}{40}$

$\Rightarrow k=\frac{1}{4}$