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Mathematics

What point on x-axis is equidistant from the points (6, 7) and (4, -3) ?

Section Formula

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Answer

We know that,

y-coordinate on x-axis = 0.

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

Distance formula = (x2x1)2+(y2y1)2\sqrt{(x2 - x1)^2 + (y2 - y1)^2}

Since, (x, 0) is equidistant from (6, 7) and (4, -3)

(x6)2+(07)2=(x4)2+[0(3)]2x2+3612x+(7)2=x2+168x+32x2+3612x+49=x2+168x+9\therefore \sqrt{(x - 6)^2 + (0 - 7)^2} = \sqrt{(x - 4)^2 + [0 - (-3)]^2} \\[1em] \Rightarrow \sqrt{x^2 + 36 - 12x + (-7)^2} = \sqrt{x^2 + 16 - 8x + 3^2} \\[1em] \Rightarrow \sqrt{x^2 + 36 - 12x + 49} = \sqrt{x^2 + 16 - 8x + 9}

Squaring both sides we get :

x212x+85=x28x+25x2x28x+12x=85254x=60x=15.\Rightarrow x^2 - 12x + 85 = x^2 - 8x + 25 \\[1em] \Rightarrow x^2 - x^2 - 8x + 12x = 85 - 25 \\[1em] \Rightarrow 4x = 60 \\[1em] \Rightarrow x = 15.

(x, 0) = (15, 0).

Hence, (15, 0) is equidistant from (6, 7) and (4, -3).

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