Question

Let $PAB$ be a triangle in which $PA=PB$ and the coordinates of $A$ and $B$ are $(-1,-6)$ and $(-5,-2)$ respectively. If the area of $△PAB$ is $8$ square units, then the coordinates of $P$ are

• A. $(-1,-2)$
• B. $(-2,-3)$
• C. $(-4,-5)$
• D. $(-5,-6)$

Answer 1

Answer: (A), (D)

The correct options are
A $(-1,-2)$
D $(-5,-6)$

Let the coordinates of $P$ be $(h,k)$
Now, $PA=PB$
$\Rightarrow P{A}^2=P{B}^2\Rightarrow (h+1{)}^2+(k+6{)}^2=(h+5{)}^2+(k+2{)}^2\Rightarrow 2h+1+12k+36=10h+25+4k+4\Rightarrow -8h+8k+8=0\Rightarrow h=k+1$

Now, the area of the $△PAB$
$8=\frac{1}{2}\left|\begin{array}{cccc}{x}_1& x_2& x_3& x_1\\ y_1& y_2& y_3& y_1\end{array}\right|\Rightarrow 16=\left|\begin{array}{cccc}k+1& -1& -5& k+1\\ k& -6& -2& k\end{array}\right|\Rightarrow 16=|(-6k-6+k)+(2-30)+(-5k+2k+2)|\Rightarrow 16=|-8k-32|\Rightarrow |-k-4|=2\Rightarrow -k-4=\pm 2\Rightarrow -k=4\pm 2\Rightarrow -k=2,6\therefore k=-2,-6\therefore h=-1,-5$

Hence, the coordinates of $P$ are $(-1,-2)$ and $(-5,-6)$