Answer:
Since, P us equidistant from points A and B, \[\therefore \] \[PA=PB\] Or, \[{{(PA)}^{2}}={{(PB)}^{2}}\] \[{{(a+b-x)}^{2}}+{{(b-a-y)}^{2}}={{(a-b-x)}^{2}}+{{(a+b-y)}^{2}}\] \[{{(a+b)}^{2}}+{{x}^{2}}-2ax-2bx+{{(b-a)}^{2}}+{{y}^{2}}-2by+2ay\]\[={{(a-b)}^{2}}+{{x}^{2}}-2ax+2bx+{{(a+b)}^{2}}+{{y}^{2}}-2ay-2by\] \[-2bx+2ay=2bx-2ay\] \[4ay=4bx\] \[ay=bx\] \[bx=ay\] Hence Proved.
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