Question

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

Answer 1

Formula:

Area of triangle $=\frac{1}{2}\left[x_1\right(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2\left)\right]$

If the area of a triangle is zero, then the points are collinear.

Given,

$A(a,0),B(0,b),C(1,1)$

Area of $△ABC=\frac{1}{2}\left[a\right(b-1)+0(1-0)+1(0-b\left)\right]$

$=\frac{1}{2}[ab-a+0-b]$

from given condition, we have,

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

$\Rightarrow ab=a+b$

$=(a+b)-a-b=0$sq.units

Hence the given points are collinear