मराठी

Write the coordinates of a point on X-axis which is equidistant from the points (−3, 4) and (2, 5). - Mathematics

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प्रश्न

Write the coordinates of a point on X-axis which is equidistant from the points (−3, 4) and (2, 5).

बेरीज
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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

`d =  sqrt( (x_1 - x_2 )^2 + (y_1 - y_2 )^2)`

Here we are to find out a point on the x−axis which is equidistant from both the points

A(−3, 4) and B(2, 5).

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( ( - 3- x)^2 + (4 - y )^2)`

     `= sqrt((-3-x)^2 + (4 - 0 )^2))`

`AC = sqrt((-3-x)^2 + ( 4)^2`

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

      `= sqrt((2 -x)^2 + ( 5 - 0)^2)`

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

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

                                 AC = BC 

`sqrt((-3-x)^2 + (4)^2 ) = sqrt((2 - x)^2 + (5)^2)`

Squaring on both sides we have,

`(-3-x)^2 + (4)^2 = (2 -x)^2 + (5)^2`

`9 +x^2 + 6x + 16 = 4 +x^2 - 4x + 25`

                       `10x  = 4`

                          ` x = 2/5`

Hence the point on the x-axis which lies at equal distances from the mentioned points is`(2/5 , 0).` 

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पाठ 6: Co-Ordinate Geometry - Exercise 6.6 [पृष्ठ ६२]

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आरडी शर्मा Mathematics [English] Class 10
पाठ 6 Co-Ordinate Geometry
Exercise 6.6 | Q 11 | पृष्ठ ६२

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