Share

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

ConceptConcepts of Coordinate Geometry

#### 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).

Is there an error in this question or solution?

#### Video TutorialsVIEW ALL [1]

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