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