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If the point $A( 0,2)$ is equidistant from the points $B( 3,\ p)$ and $C( p,\ 5)$, find P. Also find the length of AB.
Given: Three points $A( 0,\ 2)$, $B( 3,\ p)$ and $C( p,\ 5)$. point A is equidistant from the points B and C.
To do: To find the value of p and the length of AB.
Solution: We know if there two points $( x_{1} ,\ y_{1})$ and $( x_{2} ,\ y_{2})$,
Distance between the two points$=\sqrt{( x_{2} -x_{1})^{2} +( y_{2} -y_{1})^{2}}$
On using the distance formula,
We have,
AB=$\sqrt{( 3-0)^{2} +( p-2)^{2}}$
$\Rightarrow AB=\sqrt{9+( p-2)^{2}}$
Similarly $AC=\sqrt{( p-0)^{2} +( 5-2)^{2}}$
$\Rightarrow AC=\sqrt{p^{2} +9}$
As given in the question, A is equidistant from B and C.
$AB=AC$
$\Rightarrow \sqrt{9+( p-2)^{2}} =\sqrt{p^{2} +9}$
$\Rightarrow 9+( p-2)^{2} =p^{2} +9$
$\Rightarrow p^{2} +4-4p+9=p^{2} +9$
$\Rightarrow 4-4p=0$
$\Rightarrow 4p=4$
$\Rightarrow p=\frac{4}{4}$
$\Rightarrow p=1$
On subtituting $p=1$,
We have $AB=\sqrt{9+( 1-2)^{2}}$
$AB=\sqrt{9+1}$
$\Rightarrow AB=\sqrt{10} \ units$
Thus the value of $p=1$ and length of $AB=\sqrt{10} \ units$.
Updated on: 10-Oct-2022
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