The locus of the point which is equidistant f | vedclass

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(B) Let $(x, y)$ be the point which is equidistant from the point $(1, 1)$ and the line $x+y+1=0$. By the definition of distance,the distance from $(x

Vedclass · Accepted Answer

$(x-y)^2-6(x+y)+3=0$

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(B) Let $(x, y)$ be the point which is equidistant from the point $(1, 1)$ and the line $x+y+1=0$. By the definition of distance,the distance from $(x, y)$ to $(1, 1)$ is $\sqrt{(x-1)^2+(y-1)^2}$. The distance from $(x, y)$ to the line $x+y+1=0$ is $\frac{|x+y+1|}{\sqrt{1^2+1^2}} = \frac{|x+y+1|}{\sqrt{2}}$. Equating the two distances: $\sqrt{(x-1)^2+(y-1)^2} = \frac{|x+y+1|}{\sqrt{2}}$ Squaring both sides: $(x-1)^2+(y-1)^2 = \frac{(x+y+1)^2}{2}$ $2(x^2-2x+1+y^2-2y+1) = x^2+y^2+1+2xy+2x+2y$ $2x^2+2y^2-4x-4y+4 = x^2+y^2+2xy+2x+2y+1$ $x^2+y^2-2xy-6x-6y+3 = 0$ $(x-y)^2-6(x+y)+3 = 0$.

The locus of the point which is equidistant from the point $(1,1)$ and the line $x+y+1=0$ is

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