Find a point on the x-axis, which is equidistant from the points (7, 6) and (3, 4).
Find a point on the x-axis, which is equidistant from the points (7, 6) and (3, 4).
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Let (x, 0) be the points on the x-axis.
Then the distance will be same;
(x – 3)2 + 16 = (x – 7)2 + 36
⇒ x2 – 6x + 9 + 16 = x2 – 14x + 49 + 36
⇒ 14x – 6x = 49 + 36 – 9 – 16
⇒ 8x = 60 ⇒ x = \(\frac{15}{2}\)
Hence the point is (\(\frac{15}{2}\), 0).
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