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

Coordinate Geometry

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Answer

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

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

Distance of the point (0, a) from the points (7, 6) and (-3, 4) is equal.

⇒(7−0)2+(6−a)2=(−3−0)2+(4−a)2⇒72+(6−a)2=(−3)2+(4−a)2⇒49+36+a2−12a=9+16+a2−8a⇒9+16+a2−8a−49−36−a2+12a=0⇒4a−60=0⇒4a=60⇒a=604⇒a=15\Rightarrow \sqrt{(7 - 0)^2 + (6 - a)^2} = \sqrt{(-3 - 0)^2 + (4 - a)^2}\\[1em] \Rightarrow 7^2 + (6 - a)^2 = (-3)^2 + (4 - a)^2\\[1em] \Rightarrow 49 + 36 + a^2 - 12a = 9 + 16 + a^2 - 8a\\[1em] \Rightarrow 9 + 16 + a^2 - 8a - 49 - 36 - a^2 + 12a = 0\\[1em] \Rightarrow 4a - 60 = 0\\[1em] \Rightarrow 4a = 60\\[1em] \Rightarrow a = \dfrac{60}{4}\\[1em] \Rightarrow a = 15

Hence, the point on y-axis that is equidistant from (7, 6) and (-3, 4) is (0, 15).

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