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Find a Point on Y-axis Which is Equidistant from the Points (5, -2) and (-3, 2). - CBSE Class 10 - Mathematics

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Question

Find a point on y-axis which is equidistant from the points (5, -2) and (-3, 2).

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(5,2) and B(3,2).

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((5 - x)^2 + (-2 - y)^2)`

`= sqrt((5 - 0)^2 + (-2 - y)^2)`

`AC = sqrt((5)^2 + (-2-y)^2)`

`BC = sqrt((-3-x)^2 + (2 - y)^2)`

`= sqrt((-3-0)^2 + (2 - y)^2)`

`BC = sqrt((-3)^2 + (2 - y)^2)`

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

AC = BC

`sqrt((5)^2 + (-2-y)^2) = sqrt((-3)^2 + (2 - y)^2)`

Squaring on both sides we have,

`(5)^2 + (-2 -y)^2 = (-3)^2 + (2 - y)^2 `

`25 + 4 + y^2 + 4y = 9 + 4 + y^2 - 4y` 

8y = -16

y = -2

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

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Solution Find a Point on Y-axis Which is Equidistant from the Points (5, -2) and (-3, 2). Concept: Concepts of Coordinate Geometry.
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