Question

Consider the point $A\equiv (0,1)$ and $B\equiv (2,0).{}^{\prime }{P}^{\prime }$ be a point on the line $4x+3y+9=0.$ Coordinate of the point $P$ such that $|PA-PB|$ is maximum, is

• A. $(-\frac{12}{5},\frac{17}{5})$
• B. $(-\frac{84}{5},\frac{13}{5})$
• C. $(-\frac{6}{5},\frac{17}{5})$
• D. $(-\frac{24}{5},\frac{17}{5})$

Answer 1

The correct option is D $(-\frac{24}{5},\frac{17}{5})$

We know that $|PA-PB|\le AB.$
Thus, for $|PA-PB|$ to be maximum, point $A,B,P$ must be collinear
Now the equation of $AB$ is
$y=-\frac{1}{2}(x-2)$
$\Rightarrow x+2y=2\cdots \left(1\right)$
Given line is $4x+3y+9=0\cdots \left(2\right)$
Solving $\left(1\right)$ and $\left(2\right)$, we get
$P\equiv (-\frac{24}{5},\frac{17}{5})$