Question

The coordinates of two points A and B are (3, 4) and (5, -2) respectively. If P is any point such that PA = PB and area of $\Delta PAB=10$ square units, then find the coordinates of point P.

• A. (1, 0) or (7, 2)
• B. (0, 1) or (2, 7)
• C. (1, 0) or (2, 7)
• D. (0, 1) or (7, 2)

Answer 1

The correct option is A (1, 0) or (7, 2)

Let the coordinates of point P be (x, y)

Given: Area of $\Delta PAB=10$


$\Rightarrow \frac{1}{2}\left|x\right(4-(-2))+3(-2-y)+5(y-4)|=10$

$\Rightarrow \frac{1}{2}\left|x\right(6)+3(-2-y)+5(y-4\left)\right|=10$

$\Rightarrow |6x-6-3y+5y-20|=20$

$\Rightarrow |6x+2y-26|=20$

$\Rightarrow 6x+2y-26=\pm 20$

$6x+2y-26=20$

$\Rightarrow 6x+2y-46=0$

$\Rightarrow 2(3x+y-23=0)$

$\Rightarrow 3x+y-23=0$ ...(1)

Also, $6x+2y-26=-20$

$\Rightarrow 6x+2y-6=0$

$\Rightarrow 2(3x+y-3)=0$

$\Rightarrow 3x+y-3=0$ ...(2)

Now, PA = PB

${\left(\sqrt{(x-3{)}^2+(y-4{)}^2}\right)}^2$

$={\left(\sqrt{(x-5{)}^2+(y-(-2){)}^2}\right)}^2$

$\Rightarrow (x-3{)}^2+(y-4{)}^2=(x-5{)}^2+(y+2{)}^2$

$\Rightarrow x^2+9-6x+y^2+16-8y=x^2+25-10x+y^2+4+4y$

$\Rightarrow -6x-8y=-10x+4+4y$

$\Rightarrow -6x-8y+10x-4-4y=0$

$\Rightarrow 4x-12y-4=0\Rightarrow 4(x-3y-1)=0$

$\Rightarrow x-3y-1=0$

$\Rightarrow x=3y+1$ ...(3)

Substituting $x=1+3y$ from (3) in (1), we get:

$3x+y-23=0$

$\Rightarrow 3(1+3y)+y-23=0$

$\Rightarrow 3+9y+y-23=0$

$\Rightarrow 10y-20=0\Rightarrow y=\frac{20}{10}$

$\therefore y=2$

Put the value of x in (3), $x=1+3y=1+3\times 2$

$\therefore x=7$

Substituting $x=1+3y$ from (3) in (2), we get:

$3x+y-3=0$

$\Rightarrow 3(1+3y)+y-3=0$

$\Rightarrow 3+9y+y-3=0$

$\Rightarrow 10y=0$

$\therefore y=0$

from (3), $x=1+3y=1+3\times 0$

$\therefore x=1$

Thus, coordinates of point P are (1, 0) or (7, 2)

Hence, option a is correct.