Question

Consider $A(0,1)$ and $B(2,0)$ and P be a point of the line $4x+3y+9=0$, co-ordinates of P such $|PA-PB|$ is maximum is?

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

Answer 1

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

Given equation $4x+3y+9=0$ ……..(1)

Let coordinate of P is (x,y)

Now

$PA=\sqrt{{(x-0)}^2+{(y-1)}^2}=\sqrt{x^2+{(y-1)}^2}$

and $PB=\sqrt{{(x-2)}^2+{(y-0)}^2}=\sqrt{{(x-2)}^2+y^2}$

Now using Cosine formula

We get

cosθ=$\frac{P{A}^2+P{B}^2-AB}{2.PA.PB}$

Now we know that

$|cos\theta |\le 1$ $\Rightarrow \frac{P{A}^2+P{B}^2-AB}{2.PA.PB}\le 1$

So we get

$P{A}^2+P{B}^2-A{B}^2\le 2.PA.PB$

$P{A}^2+P{B}^2-2PA.PB\le A{B}^2$

${(PA-PA)}^2\le A{B}^2$

$PA-PB\le AB$

So we get |PA−PB|≤|AB|

and equality hold when θ=π

Means A,P,B

are Collinear.

So equation of line P,A,B

is

Now,equation which pass through point A, and B

$y-1=\frac{0-1}{2-0}(x-0)$

$y-1=-\frac{1}{2}x$

$y=-\frac{1}{2}x+1$ ……..(2)

From equation (1) and (2)

$4x+3(-\frac{1}{2}x+1)+9=0$

$x=-\frac{24}{5}$

Put the value of x in (2) equation

We get $y=\frac{17}{5}$

Hence , the value of $x,y=(\frac{-24}{5},\frac{17}{5})$