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.

Given that point A is equidistant from points B and C , so AB = AC


By distance formula, as shown below:



AB = √{(8 – x)2 + (– 2 – 2)2}


= √{(8 – x)2 + (– 4)2}


= √{64 + x2 – 16x + 16}


= √{80 + x2 – 16x}


AC = √{(2 – x)2 + (– 2 – 2)2}


= √{(2 – x)2 + (4)2}


= √{4 + x2 – 4x + 16}


= √{20 + x2 – 4x}


Now, AB = AC


Squaring both sides, we get,


(80 + x2 – 16x) = (20 + x2 – 4x)


60 = 12x


x = 5


⇒ AB = √{80 + x2 – 16x}


⇒ AB = √(80 + 52 – 16× 5)


= 5 units


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