Question:
If points $(a, 0),(0, b)$ and $(1,1)$ are collinear, then $\frac{1}{a}+\frac{1}{b}=$
(a) 1
(b) 2
(c) 0
(d) −1
Solution:
We have three collinear points $\mathrm{A}(a, 0) ; \mathrm{B}(0, b) ; \mathrm{C}(1,1)$.
In general if $\mathrm{A}\left(x_{1}, y_{1}\right) ; \mathrm{B}\left(x_{2}, y_{2}\right) ; \mathrm{C}\left(x_{3}, y_{3}\right)$ are collinear then,
$x_{1}\left(y_{2}-y_{3}\right)+x_{2}\left(y_{3}-y_{1}\right)+x_{3}\left(y_{1}-y_{2}\right)=0$
So,
$a(b-1)+0(1-0)+1(0-b)=0$
So,
$a b=a+b$
Divide both the sides by $(a b)$,
$\frac{1}{a}+\frac{1}{b}=1$
So the answer is (a)