Question

If three points $(x_1,y_1),(x_2,y_2),(x_3,y_3)$ lie on the same line, prove that $\frac{y_2-y_3}{x_2{x}_3}+\frac{y_3-y_1}{x_3{x}_1}+\frac{y_1-y_2}{x_1{x}_2}=0.$

Answer 1

If the three points are collinear, then the area of triangle formed by them is zero.
Since, the sides of triangle are $(x_1,y_1),(x_2,y_2),(x_3,y_3)$
then, Area of triangle = $\frac{1}{2}[\left(x_1\right(y_2-y_3\left)\right)+\left(x_2\right(y_3-y_1\left)\right)+\left(x_3\right(y_1-y_2\left)\right)]$

$\Rightarrow$ $[\left(x_1\right(y_2-y_3\left)\right)+\left(x_2\right(y_3-y_1\left)\right)+\left(x_3\right(y_1-y_2\left)\right)]=0$

Now, divide both sides with $\left(x_1{x}_2{x}_3\right)$

We get,

$\Rightarrow$ $\frac{y_2-y_3}{x_2{x}_3}+\frac{y_3-y_1}{x_3{x}_1}+\frac{y_1-y_2}{x_1{x}_2}=0.$