Question

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.

Answer 1

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

\(d_1=\frac{\left|x+y-5\right|}{\sqrt{(1)^2+(1)^2}} \) and \(d_2=\frac{\left|3x-2y+7\right|}{\sqrt{(3)^2+(-2)^2}}\)

i.e, \(d_1=\frac{\left|x+y-5\right|}{\sqrt{2}}\)  and \(d_2=\frac{\left|3x-2y+7\right|}{\sqrt{13}}\)

It is given that \(d_1+d_2=10\)

\(∴ \frac{\left|x+y-5\right|}{√\sqrt{2}}+\frac{\left|3x-2y+7\right|}{\sqrt{13}}=10\)

\(⇒ \sqrt{13}\left|x+y-5\right|+\sqrt{2}\left|3x-2y+7\right|-10\sqrt{26}=0\)

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

[Assuming \((x+y-5)\) and \((3x-2y+7)\)  are postive]

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

\(⇒ x(\sqrt{13}+3\sqrt2)+y(\sqrt{13}-2\sqrt2)+(7\sqrt2-10\sqrt{26}-5\sqrt{13})=0\), which is the equation of a line.

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

Thus, point P must move on a line.