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Find a point on the x-axis, which is equidistant from the points (7, 6) and (3, 4). |
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Answer» 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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