A(2,0), B(0,2), C(-2,0) are three points. Let | vedclass

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(B) The equations of the lines are:$AB: x+y-2=0$$BC: x-y+2=0$$CA: y=0$Let $P(x, y)$ be a point. The perpendicular distances are:$a = \frac{|x+y-2|}{\s

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$\sqrt{2}|y|=|x-y+2|-|x+y-2|$

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(B) The equations of the lines are:$AB: x+y-2=0$$BC: x-y+2=0$$CA: y=0$Let $P(x, y)$ be a point. The perpendicular distances are:$a = \frac{|x+y-2|}{\sqrt{1^2+1^2}} = \frac{|x+y-2|}{\sqrt{2}}$$b = \frac{|x-y+2|}{\sqrt{1^2+(-1)^2}} = \frac{|x-y+2|}{\sqrt{2}}$$c = |y|$Given $a, b, c$ are in arithmetic progression,we have $2b = a+c$ or $2a = b+c$ or $2c = a+b$.Assuming the order $a, b, c$,we have $2b = a+c$:$2\frac{|x-y+2|}{\sqrt{2}} = \frac{|x+y-2|}{\sqrt{2}} + |y|$$\sqrt{2}|x-y+2| = |x+y-2| + \sqrt{2}|y|$Rearranging gives $\sqrt{2}|y| = |x-y+2| - |x+y-2|$ (assuming specific signs for the absolute values based on the region).Checking the options,option $B$ matches the derived condition.

$A(2,0), B(0,2), C(-2,0)$ are three points. Let $a, b, c$ be the perpendicular distances from a variable point $P(x, y)$ onto the lines $AB, BC$ and $CA$ respectively. If $a, b, c$ are in arithmetic progression,then the locus of $P$ is

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