What is the relation between X and Y if the point (x, y) is equidistant?

Geometry is the branch of the study of shapes and structures in mathematics. It seals with points, lines, curves, and relations between them. The relation between the various point plotted on a plane surface are determined by using the standard distance formula. By the use of the formula, we determine the distance between two points involved in the operations and the type of relationship between them. The given article demonstrates the relation between two points X and Y if the point XY is equidistant. 

Distance formula

Distance formula is used to find the distance between any two points or between a point to a line or distance between two parallel lines. It can also used to find the distance between two points present in any plane, like 2D, 3D, parallel plane, etc. The standard distance formula is

D = √(x₂ − x₁)² + (y₂ − y₁)²

What is the relation between X and Y if the point (x, y) is equidistant?

Equidistant relation is defined as determining the distance of a point from two other given points in the line or graph. The point lies at an equal distance from both the point which can be calculated by using the distance formula and coordinate of the two given points.

Let us consider the two points to be X and Y with coordinates (x1, y1) and (x2, y2) respectively. Now, derive the relation between X and Y by using points X(x1, y1) and Y(x2, y2). So, to show the relation between X and Y is equidistant. Let P(x, y) be equidistant from the points X(x1, y1) and Y(x2, y2).

So, XP = YP

Squaring on both sides, we have

=>(XP)² = (YP)²

Using distance formula,

=>√(x2 − x1)² + (y2 − y1)²

Now,

=>(XP)² = (YP)²

=>(x – x1)² + (y – y1)² = (x – x2)² + (y – y2)²

This is the relation between X and Y. Now let us discuss this with the help of an example: 

Example:

Let P(x, y) be equidistant from the points A(7, 1) and B(3, 5).

So, AP = BP

Squaring on both sides, we get

⇒(AP)² = (BP)²

Using, distance formula,  

Distance between (x1, y1) and (x2,​ y2​) = (x2 − x1)² + (y2 − y1)²…..(1)

Now,

 =>(AP)² = (BP)²

 =>(x − 7)² + (y −1)² = (x − 3)² + (y − 5)²

 =>x² + 49 −14x + y² + 1 − 2y = x² + 9 − 6x + y² + 25 − 10y

=>−14x + 50 − 2y = −6x + 34 − 10y

 =>−7x + 25 − y = −3x + 17 − 5y

=>−4x + 8 + 4y = 0

=>4x − 4y = 8

=>4(x − y) = 8

=>x − y = 2

Hence, this is the required relation between x and y. 

Sample Questions

Question 1. Find the value of ‘k’ if point (0, 2) is equidistant from (3, k) and (k, 5).

Solution:

Let the two points be A and B with coordinates (3, k) and (k, 5). And, P(0, 2) is equidistant.

Now,

For AP,

P(x, y) = P(0, 2)

 A(x1, y1) = A(3, k)

For BP,

P(x, y) = P(0, 2)

B(x1, y1) = B(k, 5)

=>(AP)² = (BP)²

=>(0 – 3)² + (2 – k)² = (0 – k)² + (2 – 5)²

=>9 + 4 – 4k + k² = k² + 9

=>13 – 4k = 9

=>-4k = 9-13

=>-4k = -4

=>k = 1

Question 2. Find the relation for the points A(3, 6) and B(-3, 4) if P(x, y) is equidistant.

Solution:

Let the two points be A and B with coordinates (3, 6) and (-3, 4). And, P(x, y) is equidistant.

Now,

For AP,

P(x1, y1) = P(x, y)

A(x2, y2) = A(3, 6)

For BP,

P(x1, y1) = P(x, y)

B(x2, y2) = B(-3, 4)

=>(AP)² = (BP)²

=>(x – 3)² + (y – 6)² = (x + 3)² + (y – 4)²

=>(x – 3)² + (x + 3)² = (y – 6)² + (y – 4)²

=>(x -3 + x + 3)(x – 3 – x – 3) = (y – 6 – y – 4)(y – 6 – y + 4)

=>(2x)(-6) = (2y – 10)(2)

=>-12x = 4y – 20

=>-12x – 4y + 20 = 0

=>3x + y – 5 = 0

Question 3. Find the point on Y-axis which is equidistant from points (2, -5) and (-2, 9).

Solution:

Let the two points be A and B with coordinates (2,-5) and (-2, 9). And, P(0, y) is equidistant.

Now,

For AP,

P(x1, y1) = P(0, y)

A(x2, y2) = A(2, -5)

For BP,

P(x1, y1) = P(0, y)

B(x2, y2) = B(-2, 9)

=>(AP)² = (BP)²

=>(2 – 0)² + (2 – y)² = (-2 – 0)² + (9 – y)²

=>(2)² + (2 – y)² = (-2)² + (9 – y)²

=>4 + 4 – 4y + y² = 4 + 81 – 18y + y²

=>8 – 4y = 85 – 18y

=>14y = 77

=>y = 77/14

=>y = 11/2