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प्रश्न
Find a point on the x-axis which is equidistant from the points (7, 6) and (−3, 4).
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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(7,6) and B(−3,4).
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((7 - x)^2 + (6 - y)^2)`
`= sqrt((7 - x)^2 + (6 - 0)^2)`
`AC = sqrt((7-x)^2 + (6)^2)`
`BC= sqrt((-3-x)^2 + (4- y)^2)`
`= sqrt((-3-x)^2 + (4 - 0)^2)`
`BC = sqrt((-3-x)^2 + (4)^2)`
We know that both these distances are the same. So equating both these we get,
AC = BC
`sqrt((7 - x)^2 + (6)^2) = sqrt((-3-x)^2 + (4)^2)`
Squaring on both sides we have,
`(7 -x)^2 + (6)^2 = (-3-x)^2+ (4)^2`
`49 + x^2 -14x + 36 = 9 + x^2 + 6x + 16`
20x = 60
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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