Question

The coordinates of two points $A$ and $B$ are $(3,4)$ and $(5,-2)$ respectively. If $P$ is a point not lying on any of the coordinates axes such that $PA=PB$ and area of $\Delta PAB=10$ then the coordinates of $P$ are

Answer 1

Given $A(3,4)$ and $(5,-2)$
Let P be $(a,b)$

Then $PA=\sqrt{(a-3{)}^3+(b-4{)}^2}$

$PB=\sqrt{(a-5{)}^2+(b+2{)}^2}$

Given $PA=PB$

$\therefore \sqrt{(a-3{)}^2+(b-4{)}^2}=\sqrt{a-5{)}^2+(b+2{)}^2}$

$\Rightarrow (a-2{)}^2+(b-4{)}^2=(a-5{)}^2+(b+1{)}^2$

$\Rightarrow$ On simplifying, we get $a-3b=1$....(1)

Now area of $△$ PAB $=10Sq$ units

$=\frac{1}{2}|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|=10$

$=|a(4+2)+3(-2-b)+3(-2-b)+5(b-4)|=20$

$=3a+b=23$...........(2)$

or

$3a+b=3$ ...........(3)

Solving (1) and (2)

we get (a,b)=(7,2)

solving (1) and (3)

we get (a,b)=(1,0)

co-ordinates of $P$ are $(7,2)$ or $(1,0)$