Question

If the point A(x, 2) is equidistant from the points B(8, − 2) and C(2, − 2), find the value of x. Also, find the length of AB.

Answer 1

As per the question
$$\begin{gathered} AB=AC \\ \Rightarrow \sqrt{{\left(x-8\right)}^2+{\left(2+2\right)}^2}=\sqrt{{\left(x-2\right)}^2+{\left(2+2\right)}^2} \end{gathered}$$
Squaring both sides, we get
$$\begin{gathered} {\left(x-8\right)}^2+4^2={\left(x-2\right)}^2+4^2 \\ \Rightarrow x^2-16x+64+16=x^2+4-4x+16 \\ \Rightarrow 16x-4x=64-4 \\ \Rightarrow x=\frac{60}{12}=5 \end{gathered}$$
Now,
$$\begin{gathered} AB=\sqrt{{\left(x-8\right)}^2+{\left(2+2\right)}^2} \\ =\sqrt{{\left(5-8\right)}^2+{\left(2+2\right)}^2}\left(\because x=2\right) \\ =\sqrt{{\left(-3\right)}^2+{\left(4\right)}^2} \\ =\sqrt{9+16}=\sqrt{25}=5 \end{gathered}$$
Hence, x = 5 and AB = 5 units.