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How to find a relation between two points?

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  • Last Updated : 01 Feb, 2022

Geometry is without a doubt one of the most significant areas of mathematics. Its importance is as great as that of arithmetic and algebra, both in our daily lives and in mathematical theory and practice. Geometry is used in mathematical theory and problems to determine the distance between two forms, the amount of space they take up, their sizes, and their placements.

Coordinate Geometry

A branch of geometry that is used to study the positions of points, using their coordinates. Not only that, coordinate geometry is also used to study the distance between two points, to calculate how a point divides a line segment joining two points in a ratio, finding areas of shapes in a cartesian plane, etc.

Coordinate Plane

Such a plane which is formed by the intersection of two lines, one of them being vertical and the other one horizontal, is called a co-ordinate plane in mathematics. It is a two- dimensional plane, with the vertical line being regarded as its y-axis and the horizontal one as the x-axis. The point of intersection of the two lines in the plane is called origin and is denoted by O. The figures on a matching grid are used to detect points. A coordinate plane can be used to graph points, lines, and much further. It acts as a chart and yields precise directions from one point to another.

How to find a relation between two points?

Solution:

In geometry, there are indefinite number of ways to develop a relation between two points, one of them being the application of distance formula.

Distance Formula

In simple terms, distance is the measurement of how far two things/objects/points are separated. Clearly, the distance formula is a mathematical equation for calculating the distance between two objects/points based on their coordinates. It should be noted that the points in question do not necessarily have to be in the same quadrant. The distance formula is used primarily in the coordinate system in mathematics to determine how far apart two points are in a coordinate plane using their coordinates, making it extremely important in the subject of geometry.

\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

Example: Let O be the starting point. If OP=OQ, we have a relationship between two points P and Q in a plane. 

Since this is true for (P,Q) as well as (Q,P) it is a symmetric relation.

Similarly, the equivalence of this relation can also be proved. 

Sample Problems

Question 1. Find a relation between x and y if the point (x, y) is equidistant from (3, 6) and (-3, 4).

Solution:

Given: Point P(x, y) is equidistant from both A(3, 6) and B(-3, 4).

Using distance formula, distance between P(x, y) and A(3, 6) is given by:

D1 = \sqrt{(3-x)^2+(6-y)^2}

Using distance formula, distance between P(x, y) and B(-3, 4) is given by:

D2 = \sqrt{(-3-x)^2+(4-y)^2}

Since it is given that D1 = D2, we have:

\sqrt{(3-x)^2+(6-y)^2} = \sqrt{(-3-x)^2+(4-y)^2}

Square both sides. Then,

x2 + 9 – 6x + y2 + 36 – 12y = x2 + 9 + 6x + y2 + 16 – 8y

⇒ 12x + 4y = 20

⇒ 3x + y = 5

⇒ 3x + y – 5 = 0

Question 2. What is the relation between X and Y if the point (x, y) is equidistant from (1, 2) and (3, 5)?

Solution:

Given: Point P(x, y) is equidistant from both A(1, 2) and B(3, 5).

Using distance formula, distance between P(x, y) and A(1, 2) is given by:

D1 = \sqrt{(1-x)^2+(2-y)^2}

Using distance formula, distance between P(x, y) and B(3, 5) is given by:

D2 = \sqrt{(3-x)^2+(5-y)^2}

Since it is given that D1 = D2, we have:

\sqrt{(1-x)^2+(2-y)^2} = \sqrt{(3-x)^2+(5-y)^2}

Square both sides. Then,

x2 + 1 – 2x + y2 + 4 – 4y = x2 + 9 – 6x + y2 + 25 – 10y

⇒ 6x – 2x + 10y – 4y = 9 – 1 + 25 – 4

⇒ 4x + 6y = 29

Question 3. Find y if the distance between the points P(2, – 3) and Q(10, y) is 10 units.

Solution:

Given: PQ = 10 units

Using distance formula, we have:

\sqrt{(2-10)^2+(-3-y)^2} = 10^2

⇒ \sqrt{(-8)^2+(y+3)^2} = 100

Square both sides.

⇒ 64 + (y +3)2 = 100

⇒ (y +3)2 = 36

⇒ y = 3 or −9

Question 4. What is the relation between X and Y if the point (x, y) is equidistant from (3, 6) and (-3, 5)?

Solution:

Given: Point P(x, y) is equidistant from both A(3, 6) and B(-3, 5).

Using distance formula, distance between P(x, y) and A(3, 6) is given by:

D1 = \sqrt{(3-x)^2+(6-y)^2}

Using distance formula, distance between P(x, y) and B(-3, 5) is given by:

D2 = \sqrt{(-3-x)^2+(5-y)^2}

Since it is given that D1 = D2, we have:

\sqrt{(3-x)^2+(6-y)^2} = \sqrt{(-3-x)^2+(5-y)^2}

Square both sides. Then,

x2 + 9 – 6x + y2 + 36 – 12y = x2 + 9 + 6x + y2 + 25 – 10y

⇒ 6x + 6x + 12y – 10y = 36 – 25

⇒ 12x + 2y = 11

Question 5. What is the relation between X and Y if the point (x, y) is equidistant from (3, 6) and (-3, 2)?

Solution:

Given: Point P(x, y) is equidistant from both A(3, 6) and B(-3, 2).

Using distance formula, distance between P(x, y) and A(3, 6) is given by:

D1 = \sqrt{(3-x)^2+(6-y)^2}

Using distance formula, distance between P(x, y) and B(-3, 2) is given by:

D2 = \sqrt{(-3-x)^2+(2-y)^2}

Since it is given that D1 = D2, we have:

\sqrt{(3-x)^2+(6-y)^2} = \sqrt{(-3-x)^2+(2-y)^2}

Square both sides. Then,

x2 + 9 – 6x + y2 + 36 – 12y = x2 + 9 + 6x + y2 + 4 – 4y

⇒ 12x – 8y = -32

⇒ 3x – 2y = -8


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