# Prove that the points (a, 0), (0, b) and

Question:

Prove that the points $(a, 0),(0, b)$ and $(1,1)$ are collinear if $\frac{1}{a}+\frac{1}{b}=1$.

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,

We know area of triangle formed 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 $\Delta=\frac{1}{2}\left|x_{1}\left(y_{2}-y_{3}\right)+x_{2}\left(y_{3}-y_{1}\right)+x_{3}\left(y_{1}-y_{2}\right)\right|$

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

The three given points are A(a,0), B(0,b) and C(1,1).

$A=\frac{1}{2}|a(b-1)+1(0-b)|$

$=\frac{1}{2}|a b-a-b|$

It is given that $\frac{1}{a}+\frac{1}{b}=1$

So we have,

$\frac{1}{a}+\frac{1}{b}=1$

$\frac{a+b}{a b}=1$

$a+b=a b$

Using this in the previously arrived equation for area we have,

$A=a b-(a+b)$

$A=a b-a b$

$A=0$

Since the area enclosed by the three points is equal to 0 , the three points need to be collinear.