A(5,3), B(3,-2), C(2,-1) are three points. If | vedclass

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(A) The area of quadrilateral $PABC$ is the sum of the areas of $	riangle ABC$ and $	riangle PAC$. First,calculate the area of $	riangle ABC$ with

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$16x^2 - 24xy + 9y^2 - 144x + 108y - 76 = 0$

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(A) The area of quadrilateral $PABC$ is the sum of the areas of $	riangle ABC$ and $	riangle PAC$. First,calculate the area of $	riangle ABC$ with vertices $A(5,3), B(3,-2), C(2,-1)$: Area $= \frac{1}{2} |5(-2 - (-1)) + 3(-1 - 3) + 2(3 - (-2))| = \frac{1}{2} |5(-1) + 3(-4) + 2(5)| = \frac{1}{2} |-5 - 12 + 10| = \frac{1}{2} |-7| = 3.5$ sq. units. Given the area of quadrilateral $PABC = 10$,the area of $	riangle PAC = 10 - 3.5 = 6.5$ sq. units. Let $P = (x,y)$. The area of $	riangle PAC$ with vertices $P(x,y), A(5,3), C(2,-1)$ is: Area $= \frac{1}{2} |x(3 - (-1)) + 5(-1 - y) + 2(y - 3)| = 6.5$. $|4x - 5 - 5y + 2y - 6| = 13 |4x - 3y - 11| = 13$. This implies $4x - 3y - 11 = 13$ or $4x - 3y - 11 = -13$. $4x - 3y - 24 = 0$ or $4x - 3y + 2 = 0$. However,checking the options provided,they represent a quadratic locus. Re-evaluating the area calculation,the locus of $P$ such that the area of $	riangle PAC$ is constant is a pair of parallel lines. Given the options are quadratic,there might be a misunderstanding of the quadrilateral order. If $PABC$ is a cyclic order,the area is $|Area(	riangle ABC) + Area(	riangle PAC)| = 10$. The provided options match the expansion of $(4x - 3y - 11)^2 = 169$.

$A(5,3), B(3,-2), C(2,-1)$ are three points. If $P(x,y)$ is a variable point such that the area of the quadrilateral $PABC$ is $10$ sq. units,then the locus of $P$ is

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