If the points (h, o), (a, b) and (o, k) lie on a line, then?

We know that when three points are collinear than area of triangle is zero.

$\Rightarrow \frac{1}{2}[h(b-k)+1(a k)]=0$

$\Rightarrow b h-h k+a k=0$

$a k+b h=h k$

Dividing both sides by hk, we get

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

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