Coordinate Geometry Question 56
Question: If the point (x, y) be equidistant from the points $ (a+b,b-a) $ and $ (a-b,a+b), $ then
[MP PET 1983, 94]
Options:
A) $ ax+by=0 $
B) $ ax-by=0 $
C) $ bx+ay=0 $
D) $ bx-ay=0 $
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Answer:
Correct Answer: D
Solution:
$ {{{ x-(a+b) }}^{2}}+{{{ y-(b-a) }}^{2}}={{{ x-(a-b) }}^{2}}+{{{ y-(a+b) }}^{2}} $
$ \Rightarrow x^{2}+{{(a+b)}^{2}}-2x(a+b)+y^{2}+{{(b-a)}^{2}}-2y(b-a) $
$ =x^{2}+{{(a-b)}^{2}}-2x(a-b)+y^{2}+{{(a+b)}^{2}}-2y(a+b) $ On simplification, we get $ bx-ay=0 $
Trick: The locus will be right bisector of the line joining the given points, therefore the line must pass through the mid-points of the given point i.e. (a, b). Obviously, the line given in option (d) passes through (a, b).
BETA
