Question

If the points $(3,3),(h,0)$ and $(0,k)$ are collinear, then the value of $\frac{1}{h}+\frac{1}{k}$ is

• A. $\frac{1}{3}$
• B. $\frac{1}{6}$
• C. $\frac{1}{9}$
• D. $\frac{1}{12}$

Answer 1

The correct option is A $\frac{1}{3}$

Let $A\equiv (3,3),B\equiv (h,0)$ and $C\equiv (0,k)$
Now points $A,B,C$ are collinear, so the area of $△ABC=0$
$\Rightarrow \frac{1}{2}\left|\begin{array}{cccc}3& h& 0& 3\\ 3& 0& k& 3\end{array}\right|=0\Rightarrow |-3h+hk-3k|=0\Rightarrow 3(h+k)=hk\therefore \frac{1}{h}+\frac{1}{k}=\frac{1}{3}$