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Let $\mathrm{A}=(1,2), \mathrm{B}=(2,1), \mathrm{C}=(-1,-1)$ be three points. If $\mathrm{P}$ is a point such that the area of the quadrilateral $\mathrm{PABC}$ is twice the area of the triangle $\mathrm{PAB}$, then the equation of the locus of $\mathrm{P}$ is
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Verified Answer
The correct answer is:
$x^2-4 x y+8 y-4=0$
Let $P(x, y)$
$$
\operatorname{Ar}(P A B C)=\frac{1}{2}\left|\begin{array}{cc}
x & y \\
1 & 2 \\
2 & 1 \\
-1 & -1 \\
x & y
\end{array}\right|
$$
$\begin{aligned} &=\frac{1}{2}|((2 x+1-2-y)-(y+4-1-x)]| \\ &=\frac{1}{2}|(3 x-2 y-4)| \\ & \operatorname{Ar}(\triangle P A B)=\frac{1}{2}\left|\begin{array}{lll}x & y & 1 \\ 1 & 2 & 1 \\ 2 & 1 & 1\end{array}\right| \\ &=\frac{1}{2}|(x+y-3)| \\ & \because \quad 2 \operatorname{Ar}(\Delta P A B)=\operatorname{Ar}(P A B C)|x+y-3|=\frac{1}{2}|(3 x-2 y-4)| \\ & \Rightarrow|2 x+2 y-6|^2=|3 x-2 y-4|^2 \\ & \Rightarrow x^2-4 x y+10 y-4=0 .\end{aligned}$
$$
\operatorname{Ar}(P A B C)=\frac{1}{2}\left|\begin{array}{cc}
x & y \\
1 & 2 \\
2 & 1 \\
-1 & -1 \\
x & y
\end{array}\right|
$$
$\begin{aligned} &=\frac{1}{2}|((2 x+1-2-y)-(y+4-1-x)]| \\ &=\frac{1}{2}|(3 x-2 y-4)| \\ & \operatorname{Ar}(\triangle P A B)=\frac{1}{2}\left|\begin{array}{lll}x & y & 1 \\ 1 & 2 & 1 \\ 2 & 1 & 1\end{array}\right| \\ &=\frac{1}{2}|(x+y-3)| \\ & \because \quad 2 \operatorname{Ar}(\Delta P A B)=\operatorname{Ar}(P A B C)|x+y-3|=\frac{1}{2}|(3 x-2 y-4)| \\ & \Rightarrow|2 x+2 y-6|^2=|3 x-2 y-4|^2 \\ & \Rightarrow x^2-4 x y+10 y-4=0 .\end{aligned}$
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