Question

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

Answer 1

The equations of given lines are
$x+y-5=0\dots \left(i\right)$ and $3x-2y+7=0\dots \left(ii\right)$

Perpendicular distance of point P(x,y) from line $x+y-4=0$

$=\left|\frac{x+y-5}{\sqrt{(1{)}^2+(1{)}^2}}\right|=\left|\frac{x+y-5}{\sqrt{2}}\right|$

Perpendicular distance of point P(x,y) from line $3x-2y+7=0$

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

It is given that

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

$\Rightarrow |\sqrt{13}(x+y-5)|+|\sqrt{2}(3x-2y+7)|=10\sqrt{26}$

When $x+y-4\ge 0$ and $3x-2y+7\ge 0$

Then $\sqrt{13}(x+y-5)+\sqrt{2}(3x-2y+7)=10\sqrt{26}$

$\Rightarrow (\sqrt{13}+3\sqrt{2})x+(\sqrt{13}-2\sqrt{2})y-5\sqrt{13}+7\sqrt{2}-10\sqrt{26}=0$

Which represents a line

Similarly

When $x+y-4\ge 0$ and $3x-2y+7<0$

$x+y-5<0and3x-2y+7\ge 0$

and $x+y-5<0$ and $3x-2y+7<0$

In all the above cases it represents a line, thus point P(x,y) must move on a line.