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If the 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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Perpendicular distance $p_1$ from $P(x, y)$ to the line $x+y-5=0$ is given by $p_1=\frac{x+y-5}{\sqrt{2}}$
Perpendicular distance $p_2$ from $P(x, y)$ to the line $3 x-2 y+7=0$ is given by $p_2=\frac{3 x-2 y+7}{\sqrt{13}}$
Now $\quad p_1+p_2=10$
$\Rightarrow \frac{x+y-5}{\sqrt{2}}+\frac{3 x-2 y+7}{\sqrt{13}}=10$
or $(\sqrt{13}+3 \sqrt{2}) x+(\sqrt{13}-2 \sqrt{2}) y$
$-5 \sqrt{13}+7 \sqrt{2}-10 \sqrt{26}=0$
Which is the equation of a straight line.
Hence P moves on a line.
Perpendicular distance $p_2$ from $P(x, y)$ to the line $3 x-2 y+7=0$ is given by $p_2=\frac{3 x-2 y+7}{\sqrt{13}}$
Now $\quad p_1+p_2=10$
$\Rightarrow \frac{x+y-5}{\sqrt{2}}+\frac{3 x-2 y+7}{\sqrt{13}}=10$
or $(\sqrt{13}+3 \sqrt{2}) x+(\sqrt{13}-2 \sqrt{2}) y$
$-5 \sqrt{13}+7 \sqrt{2}-10 \sqrt{26}=0$
Which is the equation of a straight line.
Hence P moves on a line.
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