Let A=(1, 2),B=(2, 1),and C=(-1, -1) be three | vedclass

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(D) Let $P = (x, y)$. The area of quadrilateral $PABC$ is given by the shoelace formula: $	ext{Area}(PABC) = \frac{1}{2} |(x(2) + 1(1) + 2(-1) + (-1)

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$3x^2-10xy+3y^2-2x+14y-7=0$

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(D) Let $P = (x, y)$. The area of quadrilateral $PABC$ is given by the shoelace formula: $	ext{Area}(PABC) = \frac{1}{2} |(x(2) + 1(1) + 2(-1) + (-1)y) - (y(1) + 2(2) + 1(-1) + (-1)x)| = \frac{1}{2} |(2x + 1 - 2 - y) - (y + 4 - 1 - x)| = \frac{1}{2} |3x - 2y - 4|$.The area of triangle $PAB$ is given by: $	ext{Area}(	riangle PAB) = \frac{1}{2} |x(2-1) + 1(1-y) + 2(y-2)| = \frac{1}{2} |x + 1 - y + 2y - 4| = \frac{1}{2} |x + y - 3|$.Given that $	ext{Area}(PABC) = 2 	imes 	ext{Area}(	riangle PAB)$,we have:$\frac{1}{2} |3x - 2y - 4| = 2 	imes \frac{1}{2} |x + y - 3|$$|3x - 2y - 4| = 2|x + y - 3|$Squaring both sides:$(3x - 2y - 4)^2 = 4(x + y - 3)^2$$9x^2 + 4y^2 + 16 - 12xy - 24x + 16y = 4(x^2 + y^2 + 9 + 2xy - 6x - 6y)$$9x^2 + 4y^2 + 16 - 12xy - 24x + 16y = 4x^2 + 4y^2 + 36 + 8xy - 24x - 24y$$5x^2 - 20xy + 40y - 20 = 0$Dividing by $5$,we get $x^2 - 4xy + 8y - 4 = 0$.

Let $A=(1, 2)$,$B=(2, 1)$,and $C=(-1, -1)$ be three points. If $P(x, y)$ is a point such that the area of the quadrilateral $PABC$ is twice the area of the triangle $PAB$,then the equation of the locus of $P$ is:

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