If A(1,0), B(0,-2), C(2,-1) are three fixed p | vedclass

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(C) Let the coordinates of point $P$ be $(x, y)$.The area of a triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is given by $\frac{1}{2} |x

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$x^2-2xy-2x+2y+1=0$

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(C) Let the coordinates of point $P$ be $(x, y)$.The area of a triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is given by $\frac{1}{2} |x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2)|$.For $	riangle PAB$ with vertices $P(x, y), A(1, 0), B(0, -2)$:Area $= \frac{1}{2} |x(0 - (-2)) + 1(-2 - y) + 0(y - 0)| = \frac{1}{2} |2x - 2 - y|$.For $	riangle PAC$ with vertices $P(x, y), A(1, 0), C(2, -1)$:Area $= \frac{1}{2} |x(0 - (-1)) + 1(-1 - y) + 2(y - 0)| = \frac{1}{2} |x - 1 - y + 2y| = \frac{1}{2} |x + y - 1|$.Given that Area of $	riangle PAB = 	ext{Area of } 	riangle PAC$,we have:$|2x - y - 2| = |x + y - 1|$.This implies $2x - y - 2 = x + y - 1$ or $2x - y - 2 = -(x + y - 1)$.Case $1$: $x - 2y - 1 = 0$.Case $2$: $2x - y - 2 = -x - y + 1 \implies 3x = 3 \implies x = 1$.Since the options provided are quadratic,we check the product of these lines: $(x - 2y - 1)(x - 1) = x^2 - x - 2xy + 2y - x + 1 = x^2 - 2xy - 2x + 2y + 1 = 0$.

If $A(1,0), B(0,-2), C(2,-1)$ are three fixed points,then the equation of the locus of a point $P(x,y)$ such that the area of $\triangle PAB$ is equal to the area of $\triangle PAC$ is:

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