Question

Show that the points $(3,3),(h,0)$ and $(0,k)$ are collinear, if $\frac{1}{n}+\frac{1}{k}=\frac{1}{3}$.

Answer 1

For collinearity

$\left|\begin{array}{ccc}1& 3& 3\\ 1& h& 0\\ 1& 0& k\end{array}\right|=0$

This implies that,

$[1\left(hk\right)-3\left(k\right)+3\left(-h\right)=0$

$hk-3k-3h=0$

$hk=3h+3k$

$1=\frac{3}{k}+\frac{3}{h}$

$\frac{1}{h}+\frac{1}{k}=\frac{1}{3}$

Hence, these points are collinear.