Q. 74.0( 283 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(x_{1}, y_{1}) and B(x_{2}, y_{2}) is

AB = √((x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{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

⇒ x^{2} + 4 – 4x + 25 = x^{2} + 4 + 4x + 81

⇒ ( x^{2} - x^{2} ) - 4x - 4x = 81 - 25

⇒ - 8 x = 56

⇒ 8x = -56

⇒ x = - 7

Therefore, the point is (- 7, 0)

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PREVIOUSName the type of quadrilateral formed, if any, by the following points, and give reasons for your answer:
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