If the three points A

For the lines to be in a line, the slope of the adjacent lines should be the same.

Given points are $A(h, k), B\left(x_{1}, y_{1}\right)$ and $C\left(x_{2}, y_{2}\right)$

So slope of AB = BC = CA

slope $=\left(\frac{\mathrm{y}_{2}-\mathrm{y}_{1}}{\mathrm{x}_{2}-\mathrm{x}_{1}}\right)$

Slope of $A B=\left(\frac{y_{1}-k}{x_{1}-h}\right)$

Slope of $B C=\left(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\right)$

Slope of $C A=\left(\frac{y_{2}-k}{x_{2}-h}\right)$

$\Rightarrow\left(\frac{y_{1}-k}{x_{1}-h}\right)=\left(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\right)=\left(\frac{y_{2}-k}{x_{2}-h}\right)$

Now Cross multiplying the first two equality

$\left(\mathrm{y}_{1}-\mathrm{k}\right)\left(\mathrm{x}_{2}-\mathrm{x}_{1}\right)=\left(\mathrm{x}_{1}-\mathrm{h}\right)\left(\mathrm{y}_{2}-\mathrm{y}_{1}\right)$

$\Rightarrow\left(h-x_{1}\right)\left(y_{2}-y_{1}\right)=\left(k-y_{1}\right)\left(x_{2}-x_{1}\right)$

Hence proved.