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# If the sum of distances of a point from two perp[endicular lines in a plane is 1, the find the locus of the point.

Jitender Singh IIT Delhi
7 years ago
Ans: Line
Sol:
Let the coordinates of the point be (h, k). The equation of two perpendicular lines:
$ax+by+c=0$….......(1)
$-bx+ay+c^{'}=0$…...........(2)
Whera a, b, c & care constants. Let distance of point form (1) & (2) be d1& d2.
$d_{1}+d_{2}=1$
$d_{1}=\frac{|ah+bk+c|}{\sqrt{a^{2}+b^{2}}}$
$d_{2}=\frac{|-bh+ak+c^{'}|}{\sqrt{a^{2}+b^{2}}}$
$\frac{|ah+bk+c|}{\sqrt{a^{2}+b^{2}}}+\frac{|-bh+ak+c^{'}|}{\sqrt{a^{2}+b^{2}}} = 1$
$|ah+bk+c|+|-bh+ak+c^{'}| = \sqrt{a^{2}+b^{2}}$
$(a-b)h+(a+b)k+c+c^{'}=\sqrt{a^{2}+b^{2}}$
$h\rightarrow x, b\rightarrow y$
$(a-b)x+(a+b)y+c+c^{'}=\sqrt{a^{2}+b^{2}}$
There would be more solutions but all will be straight line.
Thanks & Regards
Jitender Singh
IIT Delhi