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
$\Rightarrow PA^2 = PB^2= PC^2$
$\Rightarrow PA^2 = PB^2$ and $PB^2 = PC^2$
$PA^2 = PB^2$
$\Rightarrow (5 - x)^2+ (3 - y)^2 = (5 - x)^2 + (-5 - y)^2$
$25 + x^2- 10x + 9 + y^2- 6y = 25 + x^2- 10x + 25 + y^2 + 10y$
$-6y - 10y = 25 - 9$
$\Rightarrow -16y = 16$
$y = -1$
and $PB^2 = PC^2$
$\Rightarrow (5 - x)^2 + (-5 - y)^2$
$ = (1 - x) + (-5 - y)^2$
$25 + x^2 - 10x + 25 + y^2 + 10y$
$= 1 + x^2 - 2x + 25 + y^2 + 10y$
$-10x + 2x = -24$
$ \Rightarrow -8x = -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
$\Rightarrow PA^2 = PB^2= PC^2$
$\Rightarrow PA^2 = PB^2$ and $PB^2 = PC^2$
$PA^2 = PB^2$
$\Rightarrow (5 - x)^2+ (3 - y)^2 = (5 - x)^2 + (-5 - y)^2$
$25 + x^2- 10x + 9 + y^2- 6y = 25 + x^2- 10x + 25 + y^2 + 10y$
$-6y - 10y = 25 - 9$
$\Rightarrow -16y = 16$
$y = -1$
and $PB^2 = PC^2$
$\Rightarrow (5 - x)^2 + (-5 - y)^2$
$ = (1 - x) + (-5 - y)^2$
$25 + x^2 - 10x + 25 + y^2 + 10y$
$= 1 + x^2 - 2x + 25 + y^2 + 10y$
$-10x + 2x = -24$
$ \Rightarrow -8x = -24$
$\text{x}=\frac{-24}{-8}=3$
Hence, the point $P$ is $(3, -1)$
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