If sum of the perpendicular distances of a variable point

Solution:

The equations of the given lines are

x + y – 5 = 0 … (1)

3x – 2y + 7 = 0 … (2)

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

$d_{1}=\frac{|x+y-5|}{\sqrt{(1)^{2}+(1)^{2}}}$ and $d_{2}=\frac{|3 x-2 y+7|}{\sqrt{(3)^{2}+(-2)^{2}}}$

i.e., $d_{1}=\frac{|x+y-5|}{\sqrt{2}}$ and $d_{2}=\frac{|3 x-2 y+7|}{\sqrt{13}}$

It is given that $d_{1}+d_{2}=10$.

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

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

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

$[$ Assuming $(x+y-5)$ and $(3 x-2 y+7)$ are positive $]$

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

$\Rightarrow x(\sqrt{13}+3 \sqrt{2})+y(\sqrt{13}-2 \sqrt{2})+(7 \sqrt{2}-5 \sqrt{13}-10 \sqrt{26})=0$, which is the equation of a line.

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

Thus, point P must move on a line.