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.

Explanation

Equation of given lines,
`x+y-5=0` …….`(1)`
and `3x-2y+7=0`….`(2)`
Let variable point be `P(h,k)`.
Given that,
distance of `P` from line `(1)+` distance of `P` from `(2)=10`
`implies(h+k-5)/(sqrt(1+1))+(3h-2k+7)/(sqrt(9+4))=10`
`impliessqrt(13)(h+k-5)+sqrt(2)(3h-2k+7)=10sqrt(26)`
`implies (sqrt(13)+3sqrt(2)h+(sqrt(13)-2sqrt(2))k=10sqrt(26)+5sqrt(13)-7sqrt(2)`
Now, locus of point `P` is,
`(sqrt(13)+3sqrt(2))x+(sqrt(13)-2sqrt(2))y=10(sqrt(26)+5sqrt(13)-7sqrt(2)`
Which is the equation of a line.
Therefore, `P` moves on a line.