If a point P moves such that its distances fr | vedclass

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(C) Let the coordinates of $P$ be $(x, y)$. The distance of $P(x, y)$ from $A(1, 1)$ is $\sqrt{(x-1)^2 + (y-1)^2}$.The distance of $P(x, y)$ from the

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a parabola

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(C) Let the coordinates of $P$ be $(x, y)$. The distance of $P(x, y)$ from $A(1, 1)$ is $\sqrt{(x-1)^2 + (y-1)^2}$.The distance of $P(x, y)$ from the line $x+y+2=0$ is $\frac{|x+y+2|}{\sqrt{1^2+1^2}} = \frac{|x+y+2|}{\sqrt{2}}$.According to the given condition,these distances are equal:$(x-1)^2 + (y-1)^2 = \frac{(x+y+2)^2}{2}$$2(x^2 - 2x + 1 + y^2 - 2y + 1) = x^2 + y^2 + 4 + 2xy + 4x + 4y$$2x^2 + 2y^2 - 4x - 4y + 4 = x^2 + y^2 + 2xy + 4x + 4y + 4$$x^2 + y^2 - 2xy - 8x - 8y = 0$This equation is of the form $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$,where $a=1, b=1, h=-1, g=-4, f=-4, c=0$.Here,$h^2 - ab = (-1)^2 - (1)(1) = 1 - 1 = 0$.Since $h^2 - ab = 0$,the locus represents a parabola.

If a point $P$ moves such that its distances from the point $A(1, 1)$ and the line $x+y+2=0$ are equal,then the locus of $P$ is

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