A) y=3x
B) x=y
C) x=-8y
D) -8x=y
Correct Answer: B
Solution :
Given, point \[P\left( x,\,y \right)\]is equidistant from the points A (5, 1) and B (1,5). |
So, AP = BP |
\[\Rightarrow \,\,\,A{{P}^{2}}=B{{P}^{2}}\][on squaring both sides] |
\[\Rightarrow \,\,\,{{\left( x-5 \right)}^{2}}+{{\left( y-1 \right)}^{2}}={{\left( x-1 \right)}^{2}}+{{\left( y-5 \right)}^{2}}\] |
[\[\because\] distance \[=\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}\]] |
\[\Rightarrow \,\,\,{{x}^{2}}+25-10x+{{y}^{2}}+1-2y\] |
\[={{x}^{2}}+1-2x+{{y}^{2}}+25-10y\] |
\[\left[ \because \,\,\,{{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab \right]\] |
\[\Rightarrow \,\,\,\,\,-10x+2x=-10y+2y\] |
\[\Rightarrow \,\,\,\,-8x=-8y\] |
\[\Rightarrow \,\,\,\,x=y\] |
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