A) \[ax+by=0\]
B) \[x-y=0\]
C) \[x+y=0\]
D) \[bx-ay=0\]
Correct Answer: B
Solution :
Let P be \[(x,\text{ }y)\], and we have \[A=(a+b,\,\,a-b),B=(a-b\text{, }a\text{ }+b)\] Here \[PA=PB\] \[P{{A}^{2}}=P{{B}^{2}}\] \[\Rightarrow \] \[{{[x-(a+b)]}^{2}}+{{[y-(a-b)]}^{2}}\] \[={{[x-(a-b)]}^{2}}+{{[y-(a+b)]}^{2}}\] \[\Rightarrow \] \[{{[x-(a+b)]}^{2}}-{{[x-(a-b)]}^{2}}\] \[={{[y-(a+b)]}^{2}}-{{[y-(a-b)]}^{2}}\] \[\Rightarrow [x-(a+b)+x-(a-b)]\,[x-(a+b)-x+a-b]\]\[\Rightarrow [y-(a+b)+y-(a-b)]\,[y-(a+b)-y+a-b]\] [Using \[{{x}^{2}}-{{y}^{2}}=(x+y)\,\,(x-y)\]] \[\Rightarrow \] \[(2x-2a)\,(-2b)=(2y-2a)\,(-2b)\] \[\Rightarrow \] \[2x-2a=2y-2a\] \[\Rightarrow \] \[x-y=0\]
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