How to find Equidistant Points on Y-Axis?

Distance formula is used to calculate the distance between any two points in a two-dimensional or three-dimensional plane. To find equidistant points on the y-axis we use the distance formula.

We use Pythagorean theorem to determine the required distance. Distance formula states that the distance between any two coordinates is equal to the square root of the differences between the x-coordinates and y-coordinates of the points. It is used to evaluate the distance between point to point, point to plane, and plane to plane.

The image added below shows, two points A(x, y) and B(x, y) and we will find the distance between them,

The formula to calculate the distance between two points is,

D = √((x₂ – x₁)² + (y₂ – y₁)²)

where,

• D is the distance between the points A and B
• (x₁, y₁) and (x₂, y₂) are the coordinates of A and B respectively

Consider two points A (a, b) and B (p, q) lying at a distance from each other on a two dimensional plane.

We have to find a point on the y-axis which is equidistant from these points. It is known that any point that lies on y-axis is of the form (0, y).

Suppose C is (0, y). According to the problem we can conclude that,

AC = BC

AC² = BC²

Using distance formula we have,

(0 – a)² + (y – b)² = (0 – p)² + (y – q)²

a² + y² + b² – 2yb = p² + y² + q² – 2yq

2y (q – b) = p² – q² – a² – b²

y = (p² – q² – a² – b²)/2(q – b)

The above value is calculated by substituting the given values of a, b, p and q. This gives us the point required (0, y).

Related Articles:

• Parallel Lines
• Coordinate Geometry
• Distance Between Two Points

Problem 1: Find the point on the y-axis which is equidistant from (-3, 4) and (5, 2).

Solution:

Suppose the required point is (0, y).

Using distance formula we get,

(-3 – 0)² + (4 – y)² = (5 – 0)² + (2 – y)²

9 + 16 + y² – 8y = 25 + 4 + y² – 4y

-8y + 4y – 4 = 0

4y = -4

y = -1

So, the required point is (0, -1).

Problem 2: Find the point on the y-axis which is equidistant from (6, 3) and (4, 1).

Solution:

Suppose the required point is (0, y).

Using distance formula we get,

(6 – 0)² + (3 – y)² = (4 – 0)² + (1 – y)²

36 + 9 + y² – 6y = 16 + 1 + y² – 2y

45 – 6y – 17 + 2y = 0

4y = 28

y = 7

So, the required point is (0, 7).

Problem 3: Find the point on the y-axis which is equidistant from (3, 2) and (8, 4).

Solution:

Suppose the required point is (0, y).

Using distance formula we get,

(3 – 0)² + (2 – y)² = (8 – 0)² + (4 – y)²

9 + 4 + y² – 4y = 64 + 16 + y² – 8y

13 – 4y – 80 + 8y = 0

4y = 67

y = 67/4

So, the required point is (0, 67/4).

Problem 4: Find the point on the y-axis which is equidistant from (5, 1) and (7, 2).

Solution:

Suppose the required point is (0, y).

Using distance formula we get,

(5 – 0)² + (1 – y)² = (7 – 0)² + (2 – y)²

25 + 1 + y² – 2y = 49 + 4 + y² – 4y

26 – 2y – 53 + 4y = 0

2y = 27

y = 27/2

So, the required point is (0, 27/2).

Problem 5: Find the value of x if (0, 3) is equidistant from (x, 5) and (3, 6).

Solution:

Using the distance formula we get,

(x – 0)² + (5 – 3)² = (3 – 0)² + (6 – 3)²

x² + 4 = 9 + 9

x² = 18 – 4

x² = 14

x = ±3.74

Problem 6: Find the value of x if (0, 2) is equidistant from (x, 1) and (5, 2).

Solution:

Using the distance formula we get,

(x – 0)² + (1 – 2)² = (5 – 0)² + (2 – 2)²

x² + 1 = 25 + 0

x² = 25 – 1

x² = 24

x = ±4.89

Problem 7: Find the value of x if (0, 6) is equidistant from (x, 3) and (7, 4).

Solution:

Using the distance formula we get,

(x – 0)² + (3 – 6)² = (7 – 0)² + (4 – 6)²

x² + 9 = 49 + 4

x² = 53 – 9

x² = 44

x = ±6.63

What point on the y axis is equidistant (5, 2) and (9, 2)?

The point on the y-axis that is equidistant from (- 5, 2) and (9, -2) is (0, -7).

What is the point on the y axis which is equidistant from (6, 5) and (4, 3)?

The point on the y-axis that is equidistant from (6, 5) and (4, 3) is (0, 9).

What is the point on the Y axis equidistant from (3, 4) and (7, 6)?

The point on the y-axis that is equidistant from (3, 4) and (7, 6) is (0, 15).