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.

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Solution

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

$∴ |\frac{x + y - 5}{\sqrt{1^{2} + 1^{2}}}| + |\frac{3 x - 2 y + 7}{\sqrt{3^{2} + 2^{2}}}| = 10 \Rightarrow |\frac{x + y - 5}{\sqrt{2}}| + |\frac{3 x - 2 y + 7}{\sqrt{13}}| = 10$
$(3 \sqrt{2} + \sqrt{13}) x + (\sqrt{13} - 2 \sqrt{2}) y + (7 \sqrt{2} - 5 \sqrt{13} - 10 \sqrt{26}) = 0 It is a straight line .$

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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.

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 ∴x+y-512+12+3x-2y+732+22=10⇒x+y-52+3x-2y+713=10 32+13x+13-22y+72-513-1026=0It is a straight line.