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A point $A( 0,\ 2)$ is equidistant from the points $B( 3,\ p)$ and $C( p,\ 5)$, then find the value of P.
Given: A point $A( 0,\ 2)$ is equidistant from the points $B( 3,\ p)$ and $C( p,\ 5)$.
To do: to find the value of $p$.
Solution: $\because\ A( 0,\ 2)$ is equidistant from the points $B( 3,\ p)$ and $C( p,\ 5)$.
$\therefore AB=AC$
$\Rightarrow AB^{2} =AC^{2}$
As known if there are two points $( x_{1} ,\ y_{1}) \ and\ ( x_{2} ,\ y_{2}) $,
Then the distance between them$=\sqrt{( x_{2} -x_{1})^{2} +( y_{2} -y_{1})^{2}}$
By using the distance formula,
$AB=\sqrt{( 3-0)^{2} +( p-2)^{2}}$
$=\sqrt{3^{2} +p^{2} +4-4p}$
$=\sqrt{p^{2} -4p+4+9}$
$\Rightarrow AB^{2} =p^{2} -4p+4+9$
Similarly,
$AC=\sqrt{( p-0)^{2} +( 5-2)^{2}}$
$=\sqrt{p^{2} +3^{2}}$
$=\sqrt{p^{2} +9}$
$\Rightarrow AC^{2} =p^{2} +9$
$\because \ AB^{2} =AC^{2}$
$\Rightarrow p^{2} -4p+4+9=p^{2} +9$
$\Rightarrow -4p+4=0$
$\Rightarrow 4p=4$
$\Rightarrow p=\frac{4}{4}$
$\Rightarrow p=1$
$\therefore$ The value of $p=1$.
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