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 $△APB$ is least.

• A. $(\frac{1}{3},-\frac{2}{5})$
• B. $(\frac{27}{13},-\frac{11}{13})$
• C. $(\frac{27}{7},-\frac{3}{7})$
• D. $(0,-3)$

Answer 1

The correct option is C $(\frac{27}{7},-\frac{3}{7})$

Consider the given two points.

$A=(-2,0)$ and $B=(0,4)$

Since, the equation of given line $2x-3y=9$ ......... (1)

So, in $△ABC$, $AB$ is constant.

So, for the perimeter to be minimum, $AP+BP$ must be minimum.

From AM-GM inequality,

$AP=BP$

This tells us that the point $P$ lies on the perpendicular bisector of $AB.$

Let the mid point of $AB$ be $D$, as shown in figure 1,

Slope of $AB=\frac{(4-0)}{(0+2)}=2$

Slope of $AB\times$ Slope of $DP=-1$.

$2\times m=-1$

$m=\frac{-1}{2}$

Now, the coordinate of point $D=(\frac{-2+0}{2},\frac{0+4}{2})$

$=(-1,2)$

Therefore, the equation of the line $DP$

$y-2=\frac{-1}{2}(x+1)$

$x+2y=3$ ........ (2)

From equation (1) and (2), we get

$P=(\frac{27}{7},\frac{-3}{7})$.

Hence, this is the coordinate of the required point $P$.