If sum of the perpendicular distances of a variable point $P (x, y)$ from the lines $x + y - 5$ $=$ $0$ and $3 x - 2 y + 7$ $=$ $0$ is always $10$ . Show that $P$ must move on a line.

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Solution

The perpendicular distance of $P (x, y)$ from lines (1) and (2) are respectively given by
$d_{1} = \frac{| x + y - 5 |}{(1)^{2} + (1)^{2}} and d_{2} = \frac{| 3 x - 2 y + 7 |}{\sqrt{(3)^{2} + (- 2)^{2}}}$
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 $d_{1} + d_{2} = 10$
$∴ \frac{| x + y - 5 |}{\sqrt{2}} + \frac{| 3 x - 2 y + 7 |}{\sqrt{13}} = 10$
$\Rightarrow \sqrt{13} | x + y - 5 | + \sqrt{2} | 3 x - 2 y + 7 | - 10 \sqrt{26} = 0$
$[Assuming (x+y-5) and (3x-2y+7) are positive]$
$\Rightarrow \sqrt{13} x + \sqrt{13} y - 5 \sqrt{13} + 3 \sqrt{2 x} - 2 \sqrt{2 y} + 7 \sqrt{2} - 10 \sqrt{26} = 0$
$\Rightarrow x (\sqrt{13} + 3 \sqrt{2}) + y (\sqrt{13} - 2 \sqrt{2}) + (7 \sqrt{2} - 5 \sqrt{13} - 10 \sqrt{26}) = 0$ ,
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, The point P must move on a line.

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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.

If sum of the perpendicular distances of a variable point $P (x, y)$ from the lines $x + y - 5$ $=$ $0$ and $3 x - 2 y + 7$ $=$ $0$ is always $10$ . Show that $P$ must move on a line.