Question

$A(1,4)$ and $B(3,-2)$ are two points in a plane. If point $P$ is such that $PA=PB$ and area of $\Delta PAB=40\text{sq. units,}$ then the coordinates of $P$ are

• A. $(-10,3),(14,5)$
• B. $(-10,-3),(14,5)$
• C. $(10,-3),(14,5)$
• D. $(-10,-3),(14,-5)$

Answer 1

The correct option is B $(-10,-3),(14,5)$

$\text{Let the coordinates be}P(x,y)PA=PB\Rightarrow \sqrt{(x-1{)}^2+(y-4{)}^2}=\sqrt{(x-3{)}^2+(y+2{)}^2}\Rightarrow x-3y+1=0...\left(1\right)\text{Area of}\Delta PAB=40\text{sq. unit}\frac{1}{2}\left|\begin{array}{ccc}x& y& 1\\ 1& 4& 1\\ 3& -2& 1\end{array}\right|=\pm 40\Rightarrow 6x+2y-14=\pm 803x+y+33=0...\left(2\right)\text{or,}3x+y-47=0...\left(3\right)\text{Solving eqn(1) and (2), we get}x=-10,y=-3\text{Solving eqn(1) and (3), we get}x=14,y=5\therefore P(x,y)=(-10,-3)\text{or}(14,5)$