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Find the Point on X-axis Which is Equidistant from the Points (−2, 5) and (2,−3). - Mathematics

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प्रश्न

Find the point on x-axis which is equidistant from the points (−2, 5) and (2,−3).

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उत्तर

The distance d between two points `(x_1, y_1)` and `(x_2,y_2)` is given by the formula

Here we are to find out a point on the x-axis which is equidistant from both the points A(2,5) and B(2,3)

Let this point be denoted as C(x, y).

Since the point lies on the x-axis the value of its ordinate will be 0. Or in other words, we have y = 0.

Now let us find out the distances from ‘A’ and ‘B’ to ‘C

`AC = sqrt((-2-x)^2 + (5 - y)^2)`

`= sqrt((-2 - x)^2 + (5 - 0))`

`AC = sqrt((-2-x)^2 + (5)^2)`

`BC = sqrt((2 - x)^2 + (-3-0)^2)`

`= sqrt((2 - x)^2 + (-3-0)^2)`

`BC = sqrt((2 - x)^2 + (-3)^2)`

We know that both these distances are the same. So equating both these we get,

AC = BC

`sqrt((-2-x)^2 + (5)^2) = sqrt((2 - x)^2 + (-3)^2)`

Squaring on both sides we have,

`(-2-x)^2 + (5)^2 = (2 - x)^2 + (-3)^2`

`4 + x^2 + 4x + 25 = 4 + x^2 - 4x + 9`

8x = -16

x = -2

Hence the point on the x-axis which lies at equal distances from the mentioned points is (-2, 0)

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पाठ 6: Co-Ordinate Geometry - Exercise 6.2 [पृष्ठ १६]

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आरडी शर्मा Mathematics [English] Class 10
पाठ 6 Co-Ordinate Geometry
Exercise 6.2 | Q 30 | पृष्ठ १६

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