Question

A straight line $L$ with negative slope, passes through the point $(12,3)$ and cuts the positive coordinate axes at points $P$ and $Q$. As $L$ varies, the absolute minimum value of $OP+OQ$ is $(O$ is the origin$)$

• A. $10$
• B. $18$
• C. $16$
• D. $27$

Answer 1

The correct option is D $27$

Let $m<0$ be the slope of the line $L$
Equation of line $L$ is :
$y-3=m(x-12)$
$\Rightarrow \frac{mx}{12m-3}-\frac{y}{12m-3}=1$
$\Rightarrow OP=12-\frac{3}{m},OQ=3-12m$
$OP+OQ=15+(-12m-\frac{3}{m})$
$\ge 15+2\sqrt{(-12m)\left(\frac{-3}{m}\right)}(\because AM\ge GM)$
$=15+2\times 6$
$=27$