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Question
Which point on the y-axis is equidistant from (2, 3) and (−4, 1)?
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 y-axis which is equidistant from both the points A (2, 3) and B (−4, 1).
Let this point be denoted as C(x, y).
Since the point lies on the y-axis the value of its ordinate will be 0. Or in other words, we have x = 0.
Now let us find out the distances from ‘A’ and ‘B’ to ‘C’
`AC = sqrt((2 - x)^2 + (3 - y)^2)`
`= sqrt((2 - 0)^2 + (3 - y)^2)`
`AC = sqrt((2)^2 + (3 - y)^2)`
`BC = sqrt((-4-x)^2 + (1 - y)^2)`
`= sqrt((-4-0)^2 + (1 - y)^2)`
`BC = sqrt((-4)^2 + (1 - y)^2)`
We know that both these distances are the same. So equating both these we get,
AC = BC
`sqrt((2)^2 + (3 - y)^2) = sqrt((-4)^2 + (1 - y)^2)`
`(2)^2 + (3 - y)^2 = (-4)^2 + (1 - y)^2`
`4 + 9 + y^2 - 6y = 16 + 1 + y^2 - 2y`
4y = -4
y = -1
Hence the point on the y-axis which lies at equal distances from the mentioned points is (0, -1)
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