Question
Find the coordinate of the point equidistant from three given points $A(5, 3), B(5, -5)$ and $C(1, -5).$
✓
Answer
Let the required points be $P(x, y),$ then
$PA = PB = PC.$
The points $A, B, C$ are $(5,3), (5, -5)$ and $(1, -5)$ and $(1, -5)$ recpectively
$ \Rightarrow P A^2=P B^2=P C^2$
$\Rightarrow P A^2=P B^2 \text { and } P B^2=P C^2$
$P A^2=P B^2$
$\Rightarrow(5-x)^2+(3-y)^2=(5-x)^2+(-5-y)^2$
$25+x^2-10 x+9+y^2-6 y=25+x^2-10 x+25+y^2+10 y$
$-6 y-10 y=25-9 \Rightarrow-16 y=16$
$y=-1$
$\text { and } P B^2=P C^2$
$\Rightarrow(5-x)^2+(-5-y)^2=(1-x)+(-5-y)^2$
$25+x^2-10 x+25+y^2+10 y=1+x^2-2 x+25+y^2+10 y$
$-10 x+2 x=-24 \Rightarrow-8 x=-24$
$\text{x}=\frac{-24}{-8}=3$
Hence, the point $P$ is $(3, -1)$
$PA = PB = PC.$
The points $A, B, C$ are $(5,3), (5, -5)$ and $(1, -5)$ and $(1, -5)$ recpectively
$ \Rightarrow P A^2=P B^2=P C^2$
$\Rightarrow P A^2=P B^2 \text { and } P B^2=P C^2$
$P A^2=P B^2$
$\Rightarrow(5-x)^2+(3-y)^2=(5-x)^2+(-5-y)^2$
$25+x^2-10 x+9+y^2-6 y=25+x^2-10 x+25+y^2+10 y$
$-6 y-10 y=25-9 \Rightarrow-16 y=16$
$y=-1$
$\text { and } P B^2=P C^2$
$\Rightarrow(5-x)^2+(-5-y)^2=(1-x)+(-5-y)^2$
$25+x^2-10 x+25+y^2+10 y=1+x^2-2 x+25+y^2+10 y$
$-10 x+2 x=-24 \Rightarrow-8 x=-24$
$\text{x}=\frac{-24}{-8}=3$
Hence, the point $P$ is $(3, -1)$
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