Question:medium

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 on to 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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The condition for three numbers $a, b, c$ to be in an arithmetic progression (A.P.) is $2b = a+c$. This problem combines this algebraic condition with the geometric formula for the perpendicular distance from a point to a line.

Updated On: Mar 30, 2026

$\sqrt{2}|y| = 2|x-y+2| - |x+y-2|$
$\sqrt{2}|y| = |x-y+2| - |x+y-2|$
$2|x-y+2| = \frac{|x+y-2|}{\sqrt{2}} + \frac{|x-y-2|}{\sqrt{2}}$
$2|x-y+2| = |x+(\sqrt{2}+1)y+2|$

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The Correct Option is A

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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 on to 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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The Correct Option is A

Solution and Explanation

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