Q. 74.0( 342 Votes )

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

Answer :

We have to find a point on x-axis.

Hence, its y-coordinate will be 0


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

By distance formula, Distance between two points A(x1, y1) and B(x2, y2) is

AB = √((x2 - x1)2 + (y2 - y1)2)


Distance between (x, 0) and (2, -5) = √[(x- 2)2 + (0+ 5)2]


= √[(x- 2)2 + (5)2]


Distance between (x, 0) and (-2, 9) = √[(x + 2)2 + (0 + 9)2]


= √[(x + 2)2 + (9)2]


By the given condition, these distances are equal in measure


√[(x - 2)2 + (5)2] = √[(x + 2)2 + (9)2]
Squaring both sides we get,

(x – 2)2 + 25 = (x + 2)2 + 81

⇒ x2 + 4 – 4x + 25 = x2 + 4 + 4x + 81
⇒ ( x2 - x2 ) - 4x - 4x = 81 - 25
⇒ - 8 x = 56

⇒ 8x = -56

⇒ x = - 7


Therefore, the point is (- 7, 0)

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