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If the point A(x, 2) is equidistant form the points B(8,-2) and C(2,-2), find the value of x. Also, find the value of x. Also, find the length of A B.
CBSE | Class 10 | Coordinate Geometry | Exercise 16.1 | Maths | RS Aggarwal
If the point A(x, 2) is equidistant form the points B(8,-2) and C(2,-2), find the value of x. Also, find the value of x. Also, find the length of A B.

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 \mathrm{x}=\frac{60}{12}=5

    \[\begin{aligned} &A B=\sqrt{(x-8)^{2}+(2+2)^{2}} \\ &=\sqrt{(5-8)^{2}+(2+2)^{2}} \quad(\because x=2) \end{aligned}\]

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

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

Therefore, \mathrm{x}=5 and \mathrm{AB}=5 units

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