If the point A(x, 2) is equidistant from the points B(8, − 2) and C(2, − 2),

As per the question

$A B=A C$

$\Rightarrow \sqrt{(x-8)^{2}+(2+2)^{2}}=\sqrt{(x-2)^{2}+(2+2)^{2}}$

Squaring both sides, we get

$(x-8)^{2}+4^{2}=(x-2)^{2}+4^{2}$

$\Rightarrow x^{2}-16 x+64+16=x^{2}+4-4 x+16$

$\Rightarrow 16 x-4 x=64-4$

$\Rightarrow x=\frac{60}{12}=5$

Now,

$A B=\sqrt{(x-8)^{2}+(2+2)^{2}}$

$=\sqrt{(5-8)^{2}+(2+2)^{2}} \quad(\because x=2)$

$=\sqrt{(-3)^{2}+(4)^{2}}$

$=\sqrt{9+16}=\sqrt{25}=5$

Hence, x = 5 and AB = 5 units.