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Mathematics

Find the point on the x-axis which is equidistant from (2, –5) and (–2, 9).

Coordinate Geometry

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Answer

We know that,

y-coordinate on x-axis = 0.

Let point on x-axis be (x, 0).

According to question :

Distance between (2, -5) and (x, 0) = Distance between (-2, 9) and (x, 0).

By formula,

Distance between two points (D) = (y2−y1)2+(x2−x1)2\sqrt{(y2 - y1)^2 + (x2 - x1)^2}

Substituting values we get :

⇒(x−2)2+[0−(−5)]2=[x−(−2)]2+(0−9)2⇒x2+4−4x+52=[x+2]2+(−9)2⇒x2−4x+4+25=x2+4+4x+81⇒x2−4x+29=x2+4x+85\Rightarrow \sqrt{(x - 2)^2 + [0 - (-5)]^2} = \sqrt{[x - (-2)]^2 + (0 - 9)^2} \\[1em] \Rightarrow \sqrt{x^2 + 4 - 4x + 5^2} = \sqrt{[x + 2]^2 + (-9)^2} \\[1em] \Rightarrow \sqrt{x^2 - 4x + 4 + 25} = \sqrt{x^2 + 4 + 4x + 81} \\[1em] \Rightarrow \sqrt{x^2 - 4x + 29} = \sqrt{x^2 + 4x + 85}

Squaring, both sides we get :

⇒x2−4x+29=x2+4x+85⇒x2−x2+4x+4x=29−85⇒8x=−56⇒x=−568⇒x=−7.\Rightarrow x^2 - 4x + 29 = x^2 + 4x + 85 \\[1em] \Rightarrow x^2 - x^2 + 4x + 4x = 29 - 85 \\[1em] \Rightarrow 8x = -56 \\[1em] \Rightarrow x = -\dfrac{56}{8} \\[1em] \Rightarrow x = -7.

Point = (x, 0) = (-7, 0).

Hence, point (-7, 0) is equidistant from points (2, –5) and (–2, 9).

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