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Question
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
\[\therefore \left| \frac{x + y - 5}{\sqrt{1^2 + 1^2}} \right| + \left| \frac{3x - 2y + 7}{\sqrt{3^2 + 2^2}} \right| = 10\]
\[ \Rightarrow \left| \frac{x + y - 5}{\sqrt{2}} \right| + \left| \frac{3x - 2y + 7}{\sqrt{13}} \right| = 10\]
\[\left( 3\sqrt{2} + \sqrt{13} \right)x + \left( \sqrt{13} - 2\sqrt{2} \right)y + \left( 7\sqrt{2} - 5\sqrt{13} - 10\sqrt{26} \right) = 0\]
\[\text{ It is a straight line } .\]
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