If the ratio of the distances of a variable p | vedclass

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(B) Let the coordinates of point $P$ be $(h, k)$.According to the given condition,the ratio of the distance of $P$ from $(1, 1)$ to the distance of $P

Vedclass · Accepted Answer

$3x^2+2xy+3y^2-12x-4y+4=0$

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(B) Let the coordinates of point $P$ be $(h, k)$.According to the given condition,the ratio of the distance of $P$ from $(1, 1)$ to the distance of $P$ from the line $x-y+2=0$ is $1: \sqrt{2}$.$\frac{\sqrt{(h-1)^2+(k-1)^2}}{\frac{|h-k+2|}{\sqrt{1^2+(-1)^2}}} = \frac{1}{\sqrt{2}}$$\frac{\sqrt{2} \sqrt{(h-1)^2+(k-1)^2}}{|h-k+2|} = \frac{1}{\sqrt{2}}$Squaring both sides:$\frac{2((h-1)^2+(k-1)^2)}{(h-k+2)^2} = \frac{1}{2}$$4(h^2-2h+1+k^2-2k+1) = (h-k+2)^2$$4(h^2+k^2-2h-2k+2) = h^2+k^2+4-2hk+4h-4k$$4h^2+4k^2-8h-8k+8 = h^2+k^2-2hk+4h-4k+4$$3h^2+3k^2+2hk-12h-4k+4 = 0$Replacing $(h, k)$ with $(x, y)$,the locus is $3x^2+3y^2+2xy-12x-4y+4=0$.

If the ratio of the distances of a variable point $P$ from the point $(1, 1)$ and the line $x-y+2=0$ is $1: \sqrt{2}$,then the equation of the locus of $P$ is

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