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Let $A=(4,0), B=(0,12)$ be two points in the plane. The locus of a point $C$ such that the area of triangle $A B C$ is 18 sq. units is -
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The correct answer is:
$(y+3 x-12)^{2}=81$
$\frac{1}{2}\left|\begin{array}{ccc}1 & x & y \\ 1 & 0 & 12 \\ 1 & 4 & 0\end{array}\right|=\pm 18$
$1(-48)-x(-12)+y(4)=\pm 36$
$12 x+4 y-48=\pm 36$
$3 x+y-12=\pm 9$
$(3 x+y-12)^{2}=81$
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