Question

The point $P(x,y)$ lies on the line $2x+3y+1=0$ such that $|PA-PB|$ is maximum where $A(2,0),B(0,2)$ is

• A. $(7,-5)$
• B. $(-7,5)$
• C. $(-7,-5)$
• D. None of these

Answer 1

The correct option is A $(7,-5)$

From the given equation, $y=-\frac{2x+1}{3}$.
Let $P=(x,-\frac{2x+1}{3})$.
Therefore
$P{A}^2=x^2-4x+4+\frac{4x^2+4x+1}{9}$
$P{A}^2=\frac{9x^2-36x+36+4x^2+4x+1}{9}$
$=\frac{13x^2-28x+37}{9}$ ...(i)
Similarly $P{B}^2$ $=x^2+(\frac{2x+1+6}{3}{)}^2$
$=x^2+\frac{4x^2+28x+49}{9}$
$=\frac{13x^2+28x+49}{9}$.
Now, $PA-PB$
$=f\left(x\right)$
$=\frac{1}{3}[\sqrt{13x^2-28x+37}-\sqrt{13x^2+28x+49}]$
Now, $f\left(7\right)<0$ while $f(-7)>0$
Hence, $f\left(x\right)$ is minimum at $x=7$.
Thus , $y=-\frac{2x+1}{3}$
$=-\frac{15}{3}$
$=-5$.
Hence
P=$(7,-5)$.