A(1, 0),B(0, 2),and C(1, 2) are three points | vedclass

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(B) The area of a triangle with vertices $(x_1, y_1)$,$(x_2, y_2)$,and $(x_3, y_3)$ is given by $	ext{Area} = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 -

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$4x^2 + 4xy + y^2 - 8x - 4y - 12 = 0$

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(B) The area of a triangle with vertices $(x_1, y_1)$,$(x_2, y_2)$,and $(x_3, y_3)$ is given by $	ext{Area} = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$.First,calculate the area of $	riangle ABC$ with $A(1, 0)$,$B(0, 2)$,and $C(1, 2)$:$	ext{Area}(	riangle ABC) = \frac{1}{2} |1(2 - 2) + 0(2 - 0) + 1(0 - 2)| = \frac{1}{2} |0 + 0 - 2| = 1$.Given that $	ext{Area}(	riangle PAB) = 2 	imes 	ext{Area}(	riangle ABC) = 2 	imes 1 = 2$.For $P(x, y)$,$A(1, 0)$,and $B(0, 2)$,the area is:$	ext{Area}(	riangle PAB) = \frac{1}{2} |x(0 - 2) + 1(2 - y) + 0(y - 0)| = \frac{1}{2} |-2x + 2 - y|$.Setting this equal to $2$:$\frac{1}{2} |-2x - y + 2| = 2 \Rightarrow |-2x - y + 2| = 4$.Squaring both sides:$(-2x - y + 2)^2 = 16$.$(2x + y - 2)^2 = 16$.$4x^2 + y^2 + 4 + 4xy - 8x - 4y = 16$.$4x^2 + 4xy + y^2 - 8x - 4y - 12 = 0$.

$A(1, 0)$,$B(0, 2)$,and $C(1, 2)$ are three points on the $XY$-plane. If a point $P(x, y)$ moves such that the area of $\triangle PAB$ is twice the area of $\triangle ABC$,then the locus of the point $P$ is:

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