# 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.

**Solution:**

The equations of the lines are

x + y - 5 = 0 ....(1)

3x - 2 y + 7 = 0 ....(2)

The perpendicular distance of P (x, y) from lines (1) and (2) are respectively given by

d₁ = (|x + y - 5|)/ √(1)² + (1)² and d₂ = (|3x - 2y + 7|)/ √(3)² + (- 2)²

i.e., d₁ = (|x + y - 5|)/√2 and d₂ = (|3x - 2y + 7|)/√13

It is given that d₁ + d₂ = 10

Therefore,

(|x + y - 5|)/√2 + (|3x - 2y + 7|)/√13 = 10

√13|x + y - 5| + √2|3x - 2y + 7| - 10√26 = 0

Assuming x + y - 5 = 0 and 3x - 2y + 7 = 0 are positive.

√13(x + y - 5) + √2(3x - 2y + 7) - 10√26 = 0

√13x + √13y - 5√13 + 3√2x + 7√2 - 10√26 = 0

(√13 + 3√2)x + (√13 - 2√2)y + (7√2 - 5√13 - 10√26) = 0

Since, (√13 + 3√2)x + (√13 - 2√2)y + (7√2 - 5√13 - 10√26) = 0 is the equation of a line.

Similarly, we can obtain the equation of line for any signs of x + y - 5 = 0 and 3x - 2 y + 7 = 0.

Thus, point P must move on a line

NCERT Solutions Class 11 Maths Chapter 10 Exercise ME Question 20

## 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

**Summary:**

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, then we have shown that P must move on a line