Question

If sum of 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.

Answer 1

It is given that the sum of perpendicular distances of a variable point P (x, y) from the lines x + y − 5 = 0 and 3x − 2y + 7 = 0 is always 10

$$\therefore \left|\frac{x+y-5}{\sqrt{1^2+1^2}}\right|+\left|\frac{3x-2y+7}{\sqrt{3^2+2^2}}\right|=10$$
$$\Rightarrow \left|\frac{x+y-5}{\sqrt{2}}\right|+\left|\frac{3x-2y+7}{\sqrt{13}}\right|=10$$

$$(3\sqrt{2}+\sqrt{13})x+(\sqrt{13}-2\sqrt{2})y+(7\sqrt{2}-5\sqrt{13}-10\sqrt{26})=0$$
$$\text{ It is a straight line }.$$