If three points (x1, y1) (x2, y2), (x3, y3) lie on the same line, prove that

GIVEN: If three points $\left(x_{1}, y_{1}\right),\left(x_{2} y_{2}\right)$ and $\left(x_{3}, y_{3}\right)$ lie on the same line

TO PROVE: $\frac{\left(y_{2}-y_{3}\right)}{x_{2} x_{3}}+\frac{\left(y_{3}-y_{1}\right)}{x_{3} x_{1}}+\frac{\left(y_{1}-y_{2}\right)}{x_{1} x_{2}}=0$

PROOF:

We know that three points $\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right)$ and $\left(x_{3}, y_{3}\right)$ are collinear if

$x_{1}\left(y_{2}-y_{3}\right)+x_{2}\left(y_{3}-y_{1}\right)+x_{3}\left(y_{1}-y_{2}\right)=0$

$\Rightarrow x_{1}\left(y_{2}-y_{3}\right)+x_{2}\left(y_{3}-y_{1}\right)+x_{3}\left(y_{1}-y_{2}\right)=0$

Dividing by $x_{1} x_{2} x_{3}$

$\Rightarrow \frac{x_{1}\left(y_{2}-y_{3}\right)}{x_{1} x_{2} x_{3}}+\frac{x_{2}\left(y_{3}-y_{1}\right)}{x_{1} x_{2} x_{3}}+\frac{x_{3}\left(y_{1}-y_{2}\right)}{x_{1} x_{2} x_{3}}=0$

$\Rightarrow \frac{\left(y_{2}-y_{3}\right)}{x_{2} x_{3}}+\frac{\left(y_{3}-y_{1}\right)}{x_{1} x_{3}}+\frac{\left(y_{1}-y_{2}\right)}{x_{1} x_{2}}=0$

Hence proved.