Question

Given two points $A\equiv (-2,0)$ and $B\equiv (0,4)$, then find coordinate of a point $P$ lying on the line $2x-3y=9$ so that perimeter of $\Delta APB$ is least.

• A. $\left(\frac{42}{13},-\frac{11}{3}\right)$
• B. $\left(\frac{84}{13},-\frac{74}{13}\right)$
• C. $\left(\frac{21}{17},-\frac{37}{17}\right)$
• D. $\left(0,-3\right)$

Answer 1

The correct option is A $\left(\frac{42}{13},-\frac{11}{3}\right)$

$A\equiv (-2,0)$

$B\equiv (0,4)$

$\because$ Point $P$ lies on line $2x-3y=9$

$\Rightarrow y=\frac{2x-9}{3}$

Let Point $P$ be $(\alpha ,\frac{2\alpha -9}{3})$

$\therefore$ Perimeter of $△APB\left(s\right)=\overline{AB}+\overline{AP}+\overline{BP}$

$S=\sqrt{(2{)}^2+(4{)}^2}+\sqrt{(\alpha +2{)}^2+{\left(\frac{2\alpha -9}{3}\right)}^2}+\sqrt{\alpha +{(\frac{2\alpha -9}{3}-4)}^2}$

$\Rightarrow S=2\sqrt{5}+\sqrt{(\alpha +2{)}^2+{\left(\frac{2\alpha -9}{3}\right)}^2}+\sqrt{{\alpha }^2+{\left(\frac{2\alpha -21}{3}\right)}^2}$

For perimeter to be least

$\frac{ds}{d\alpha }=0$

$\Rightarrow 0+\frac{2(\alpha +2)+\left(\frac{2\alpha -9}{3}\right).\frac{2}{3}}{2\sqrt{(\alpha +2{)}^2+{\left(\frac{2\alpha -9}{3}\right)}^2}}+\frac{2\alpha +2\left(\frac{2\alpha -21}{3}\right)\times \frac{2}{3}}{2\sqrt{{\alpha }^2+{\left(\frac{2\alpha -21}{3}\right)}^2}}=0$

Solving we get $\alpha =\frac{42}{13}$

$\therefore$ Coordinates of $P\equiv (\frac{42}{13},\frac{-11}{13})$