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If $A=(2,3)$ and $B=(-4,5)$ are two fixed points, then the locus of a point $P$ such that the area of $\triangle P A B$ is 12 square units is
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The correct answer is:
$x^2+6 x y+9 y^2-22 x-66 y-23=0$
Let the co-ordinate of the point $P$ is $\left(x_1, y_1\right)$. Then, the area of the triangle $=12$
$\Rightarrow \frac{1}{2}\left|\begin{array}{ccc}2 & 3 & 1 \\ -4 & 5 & 1 \\ x_1 & y_1 & 1\end{array}\right|=12$
$\begin{aligned} & \Rightarrow\left|2\left(5-y_1\right)-3\left(-4-x_1\right)+1\left(-4 y_1-5 x_1\right)\right|=24 \\ & \Rightarrow\left|10-2 y_1+12+3 x_1-4 y_1-5 x_1\right|=24 \\ & \Rightarrow\left|-2 x_1-6 y_1+22\right|=24 \\ & \Rightarrow 2 x_1+6 y_1-22= \pm 24 \\ & \Rightarrow x_1+3 y_1-11= \pm 12 \\ & \Rightarrow\left(x_1+3 y_1-23\right)=0 \text { or }\left(x_1+3 y_1+1\right)=0\end{aligned}$
Hence, the locus is :-
$\begin{aligned} & (x+3 y-23)(x+3 y+1)=0 \\ & \Rightarrow x^2+6 x y+9 y^2-22 x-66 y-23=0\end{aligned}$
$\Rightarrow \frac{1}{2}\left|\begin{array}{ccc}2 & 3 & 1 \\ -4 & 5 & 1 \\ x_1 & y_1 & 1\end{array}\right|=12$
$\begin{aligned} & \Rightarrow\left|2\left(5-y_1\right)-3\left(-4-x_1\right)+1\left(-4 y_1-5 x_1\right)\right|=24 \\ & \Rightarrow\left|10-2 y_1+12+3 x_1-4 y_1-5 x_1\right|=24 \\ & \Rightarrow\left|-2 x_1-6 y_1+22\right|=24 \\ & \Rightarrow 2 x_1+6 y_1-22= \pm 24 \\ & \Rightarrow x_1+3 y_1-11= \pm 12 \\ & \Rightarrow\left(x_1+3 y_1-23\right)=0 \text { or }\left(x_1+3 y_1+1\right)=0\end{aligned}$
Hence, the locus is :-
$\begin{aligned} & (x+3 y-23)(x+3 y+1)=0 \\ & \Rightarrow x^2+6 x y+9 y^2-22 x-66 y-23=0\end{aligned}$
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