A) \[ax+by=0\]
B) \[ax-by=0\]
C) \[bx+ay=0\]
D) \[bx-ay=0\]
Correct Answer: D
Solution :
\[{{\left\{ x-(a+b) \right\}}^{2}}+{{\left\{ y-(b-a) \right\}}^{2}}={{\left\{ x-(a-b) \right\}}^{2}}+{{\left\{ y-(a+b) \right\}}^{2}}\] \[\Rightarrow \,\,{{x}^{2}}+{{(a+b)}^{2}}-2x\,(a+b)+{{y}^{2}}+{{(b-a)}^{2}}-2y(b-a)\] \[={{x}^{2}}+{{(a-b)}^{2}}-2x(a-b)+{{y}^{2}}+{{(a+b)}^{2}}-2y(a+b)\] 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).
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