Question:
If the point A(0, 2) is equidistant from the points B(3, p) and C(p, 5), find the value of p. Also, find the length of AB.
Solution:
As per the question
$A B=A C$
$\Rightarrow \sqrt{(0-3)^{2}+(2-p)^{2}}=\sqrt{(0-p)^{2}+(2-5)^{2}}$
$\Rightarrow \sqrt{(-3)^{2}+(2-p)^{2}}=\sqrt{(-p)^{2}+(-3)^{2}}$
Squaring both sides, we get
$(-3)^{2}+(2-p)^{2}=(-p)^{2}+(-3)^{2}$
$\Rightarrow 9+4+p^{2}-4 p=p^{2}+9$
$\Rightarrow 4 p=4$
$\Rightarrow p=1$
Now,
$A B=\sqrt{(0-3)^{2}+(2-p)^{2}}$
$=\sqrt{(-3)^{2}+(2-1)^{2}} \quad(\because p=1)$
$=\sqrt{9+1}$
$=\sqrt{10}$ units
Hence, $p=1$ and $A B=\sqrt{10}$ units.