Find the point on x-axis which …

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Sia ? 6 years, 4 months ago
Let A(x,o) be any point on the X-axis , which is equidistant from points (-1,0) and (5,0).
{tex}\Rightarrow (x+1)^2=(x-5)^2{/tex}
⇒ x2 + 2x + 1 = x2 - 10x + 25
⇒ 2x + 1 = -10x + 25
⇒ 2x + 10x = 25 - 1
⇒ 12x = 24
⇒ x = 24/12
{tex}\Rightarrow x= 2{/tex}
Therefore , Required point is (2,0).
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