Question

$A(3,4)$ and $B(5,-2)$ are two points and P is a point such that PA$=$PB. If the area of triangle PAB is $10$ square unit, what are the coordinates of P?

• A. $(1,0)$ only
• B. $(7,2)$ only
• C. $(1,0)$ or $(7,2)$
• D. Neither $(1,0)$ nor $(7,2)$

Answer 1

Answer: (C)

$(1,0)$ or $(7,2)$

let us consider the coordinates of P is (a,b) then,

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

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

WKT PA=PB

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

On squaring both sides we get,

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

By simplifying this we get,

$\Rightarrow a-3b=1.....\left(1\right)$

Let area of triangle PAB = 10 sq.units.

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

$\Rightarrow \left|a\right(4+2)+3(-2-b)+5(b-4\left)\right|=20$

By simplifying this we get,

$3a+b=23.....\left(2\right)$

or

$3a+b=3.....\left(3\right)$

Solution form (1) and (2)

Multiply eqn (2) by 3 we get

$9a+3b=\mathrm{69.....}\left(4\right)$

Now add (1) and (4)

$10a=70$

$a=7$

substitute a value in (1)

$7-3b=1$

$b=2$

Hence $(a,b)=(7,2)$

Solution form (1) and (3)

Multiply eqn (3) by 3 we get

$9a+3b=9.....\left(5\right)$

Now add (1) and (5)

$10a=10$

$a=1$

substitute a value in (1)

$1-3b=1$

$b=0$

Hence $(a,b)=(1,0)$

Finally the coordinates of P are $(1,0)$ or $(7,2).$