**Question:**

Prove that the points (a, b), (a1, b1) and (a −a1, b −b1) are collinear if ab1 = a1b.

**Solution:**

The formula for the area ' $A$ ' encompassed by three points $\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right)$ and $\left(x_{3}, y_{3}\right)$ is given by the formula,

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

If three points are collinear the area encompassed by them is equal to 0.

The three given points are $(a, b),\left(a_{1}, b_{1}\right)$ and $\left(a-a_{1}, b-b_{1}\right)$. If they are collinear then the area enclosed by them should be 0 .

$\Delta=\frac{1}{2}\left|\left(a b_{1}+a_{1}\left(b-b_{1}\right)+\left(a-a_{1}\right) b\right)-\left(a_{1} b+\left(a-a_{1}\right) b_{1}+a\left(b-b_{1}\right)\right)\right|$

$0=\frac{1}{2}\left|\left(a b_{1}+a_{1} b-a_{1} b_{1}+a b-a_{1} b\right)-\left(a_{1} b+a b_{1}-a_{1} b_{1}+a b-a b_{1}\right)\right|$

$0=\frac{1}{2}\left|a b_{1}+a_{1} b-a_{1} b_{1}+a b-a_{1} b-a_{1} b-a b_{1}+a_{1} b_{1}-a b+a b_{1}\right|$

$0=a b_{1}-a_{1} b$

$a b_{1}=a_{1} b$

Hence we have proved that for the given conditions to be satisfied we need to have $a_{1} b=a b_{1}$.