Q.If sum of the perpendicular distances of a variable point $P(x, y)$ from the lines $x + y - 5 = 0$ and $3x - 2y + 7 = 0$ is always $10$. Show that $P$ must move on a line.

Concept: Distance from a point to a line.

The perpendicular distance of $P(x, y)$ from $x + y - 5 = 0$ is

$\frac{|x + y - 5|}{\sqrt{1^2 + 1^2}} = \frac{|x + y - 5|}{\sqrt{2}}$.

The distance from $3x - 2y + 7 = 0$ is

$\frac{|3x - 2y + 7|}{\sqrt{3^2 + (-2)^2}} = \frac{|3x - 2y + 7|}{\sqrt{13}}$.

Given that their sum is always $10$:

$\frac{|x + y - 5|}{\sqrt{2}} + \frac{|3x - 2y + 7|}{\sqrt{13}} = 10.$ …

The key idea is that the distance from a point $(x, y)$ to a line $ax + by + c = 0$ is given by $\frac{|ax + by + c|}{\sqrt{a^2 + b^2}}$. When we add two such distances and set the sum equal to a constant, we get an equation involving absolute values. The absolute values can be removed by considering the possible sign combinations of the expressions inside them. Each sign combination leads to a linear equation, meaning the point lies on a straight line.

Let’s work through it.

1. Write the distance formulas. For line $L_1: x + y - 5 = 0$, the distance from $P(x, y)$ is

For line $L_2: 3x - 2y + 7 = 0$, the distance is

2. Set up the given condition. The sum is always $10$:

3. Remove the absolute values by considering sign cases.

   Each absolute value can be either positive or zero (the expression inside is $\geq 0$) or negative (the expression inside is $< 0$). So there are four possible sign combinations for the two expressions $A = x + y - 5$ and $B = 3x - 2y + 7$. However, the problem states the sum is always 10, meaning the equation must hold for all points $P$ that satisfy it. This forces a specific relationship between $A$ and $B$ that eliminates the absolute values in a consistent way.

   Let’s denote $u = x + y - 5$ and $v = 3x - 2y + 7$. Then the equation is

For this to represent a line (not a region), the signs of $u$ and $v$ must be such that the equation becomes linear. That happens when $u$ and $v$ have the same sign (both non-negative or both non-positive), because then $|u| = \pm u$ and $|v| = \pm v$ with the same sign, giving a linear combination.

4. Consider one sign case to demonstrate. Suppose $u \geq 0$ and $v \geq 0$. Then $|u| = u$, $|v| = v$, and the equation becomes

Multiply through by $\sqrt{2}\sqrt{13}$ to clear denominators:

Expand:

Group $x$ and $y$ terms: …