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Question
Which point on the x-axis is equidistant from (5, 9) and (−4, 6)?
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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
`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 (5, 9) and B (−4, 6)
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((5 - x)^2 + (9 - y)^2)`
`= sqrt((5 - x)^2 + (9 - 0)^2)`
`AC = sqrt((5 - x)^2 + (9)^2)`
`BC = sqrt((-4 - x)^2 + (6 - y)^2)`
`= sqrt((4 + x)^2 + (6 - 0)^2)`
`BC = sqrt((4 + x)^2 + (6)^2)`
We know that both these distances are the same. So equating both these we get,
AC = BC
`sqrt((5 -x)^2 + (9)^2) = sqrt((4 + x)^2 + (6)^2)`
Squaring on both sides we have,
`(5 - x)^2 + (9)^2 = (4 + x)^2 + (6)^2`
`25 + x^2 - 10x + 81 = 16 + x^2 + 8x + 36`
18x = 54
x = 3
Hence the point on the x-axis which lies at equal distances from the mentioned points is (3,0)
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