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Home  >>  CBSE XI  >>  Math  >>  Straight Lines
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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

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  • The perpendicular distance between of a line from a point $(x_1, y_1) $ is $ d = \bigg| \large\frac{Ax_1+By_1+c}{\sqrt{A^2+B^2}} \bigg|$
The equation of the lines given are
$ \qquad x+y-5=0$--------(1)
$ \qquad 3x-2y+7=0$------(2)
The perpendicular distance of line (1) from the point $p(x,y) is
$ d_1 = \bigg| \large\frac{x+y-5}{\sqrt{1^2+1^2}} \bigg|$
$ = \large\frac{|x+y-5|}{\sqrt 2}$
Similarly the perpendicular distance of line (2) from point $p(x,y)$ is
$ d_2 = \bigg| \large\frac{3x-2y+7}{\sqrt{(3)^2+(-2)^2}} \bigg|$
$ = \large\frac{|3x-2y+7|}{\sqrt{13}}$
It is given that sum of the perpendicular distance is 10.
(i.e.,) $d_1+d_2=10$
(i.e., )$ \large\frac{|x+y-5|}{\sqrt 2}$$+ \large\frac{|3x-2y+7|}{\sqrt{13}}$$ = 10$
(i.e., ) $ \sqrt{13}\: |x+y-5|+\sqrt 2 |3x-2y+7| = 10 \sqrt{26}$
Let us take the positive values of $(x+y-5)$ and $(3x-2y+7)$
$ \sqrt{13} (x+y-5)+ \sqrt 2 (3x-2y+7)-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$
This represents the equation of a straight line.
This proves that the point $p$ must move on a line.
answered May 23, 2014 by thanvigandhi_1
 

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