Question

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

• A. $\sqrt{2}|y| = 2|x-y+2| - |x+y-2|$
• B. $\sqrt{2}|y| = |x-y+2| - |x+y-2|$
• C. $2|x-y+2| = \frac{|x+y-2|}{\sqrt{2}} + \frac{|x-y-2|}{\sqrt{2}}$
• D. $2|x-y+2| = |x+(\sqrt{2}+1)y+2|$

Hint:

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.

Answer 1

The Correct Option is A

Step 1: Equations of the lines AB, BC, and CA
Let $P(x,y)$ be the variable point. - Line AB passes through $A(2,0)$ and $B(0,2)$. Equation: \[ \frac{x}{2} + \frac{y}{2} = 1 \implies x + y - 2 = 0. \] - Line BC passes through $B(0,2)$ and $C(-2,0)$. Slope $m = \frac{0-2}{-2-0} = 1$, equation: \[ y - 2 = 1(x - 0) \implies x - y + 2 = 0. \] - Line CA passes through $C(-2,0)$ and $A(2,0)$. This is the x-axis: \[ y = 0. \]
Step 2: Expressions for perpendicular distances
The perpendicular distance from $(x_0, y_0)$ to a line $Ax+By+C=0$ is $\frac{|Ax_0+By_0+C|}{\sqrt{A^2+B^2}}$. \[ a = \text{distance from P to AB} = \frac{|x+y-2|}{\sqrt{1^2+1^2}} = \frac{|x+y-2|}{\sqrt{2}}, \] \[ b = \text{distance from P to BC} = \frac{|x-y+2|}{\sqrt{1^2+(-1)^2}} = \frac{|x-y+2|}{\sqrt{2}}, \] \[ c = \text{distance from P to CA} = \frac{|y|}{\sqrt{0^2+1^2}} = |y|. \]
Step 3: Apply the A.P. condition
For three numbers $a, b, c$ in A.P., $2b = a + c$. Substituting the distances: \[ 2 \cdot \frac{|x-y+2|}{\sqrt{2}} = \frac{|x+y-2|}{\sqrt{2}} + |y|. \]
Step 4: Simplify to find the locus
Multiply both sides by $\sqrt{2}$: \[ 2|x-y+2| = |x+y-2| + \sqrt{2}|y|. \] Rearranging: \[ \sqrt{2}|y| = 2|x-y+2| - |x+y-2|. \] \[ \boxed{\text{Locus: } \sqrt{2}|y| = 2|x-y+2| - |x+y-2|}. \]