A) \[ax=by\]
B) \[ax+by=0\]
C) \[bx+ay=0\]
D) \[bx-ay=0\]
E) \[\frac{x}{a}+\frac{y}{b}=1\]
Correct Answer: D
Solution :
Let\[P(x,\text{ }y)\]is equidistant from the points \[A(a+b,b-a)\]and\[(a-b,a+b)\] \[\therefore \] \[P{{A}^{2}}=P{{B}^{2}}\] \[\Rightarrow \] \[{{(a+b-x)}^{2}}+{{(b-a-y)}^{2}}\] \[={{(a-b-x)}^{2}}+{{(a+b-y)}^{2}}\] \[\Rightarrow \] \[bx-ay=0\]You need to login to perform this action.
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