question_answer

If the distances of P(x, y), from A(5,1) and B(\[1,\text{ }5\]) are equal, then prove that \[3x=2y\].

Answer:

Given, PA = PB

\[\Rightarrow \]            \[\sqrt{{{(x-5)}^{2}}+{{(y-1)}^{2}}}=\sqrt{{{(x+1)}^{2}}+{{(y-5)}^{2}}}\]

Squaring both sides

\[{{(x-5)}^{2}}+{{(y-1)}^{2}}={{(x+1)}^{2}}+{{(y-5)}^{2}}\]

\[\Rightarrow \]   \[{{x}^{2}}+25-10x+{{y}^{2}}+1-2y={{x}^{2}}+1+2x+{{y}^{2}}+25-10y\]

\[\Rightarrow \]               \[-10x-2y=2x-10y\]

\[\Rightarrow \]               \[-10x-2x=-10y+2y\]

\[\Rightarrow \]                         \[12x=8y\]

\[\Rightarrow \]                           \[3x=2y\]                      Hence Proved.