How to find the Trisection Points of a Line?

To find the trisection points of a line segment, you need to divide the segment into three equal parts. This involves finding the points that divide the segment into three equal lengths. In this article, we will answer “How to find the Trisection Points of a Line?” in detail including section formula.

Section Formula

Section formula is a topic that falls under coordinate geometry. It is used to find the ratio of a line segment divided by a point internally and externally. It is also used in physics to find the center of mass of systems. Mainly we study three formulas under it which are mentioned below:

• Internal Division
• External Division
• Midpoint Formula

How to find the Trisection Points of a Line?

Answer:

The formula in which a line is divided into three parts in a certain ratio of 1:2 or 2:1 internally. Use section formula to solve any problem. Section formula is mathematically given by

[Tex](\frac{mx_2+nx_1}{m+n},\frac{my_2+ny_1}{m+n})[/Tex]

Where m and n are the two integers of ratio given as m:n.

For the trisection formula, use the section formula twice,

• Step 1: Solve the points of the line segment by using the ratio m : n = 1:2.
• Step 2: Solve the points of the line segment by using the ratio m : n = 2:1.

Let’s take a look at an example, if the points are given are (3, 2) and (3, 4), according to the trisection rule, the line segment with points (3, 2) and (3, 4) will be divided into the ratios of 1:2 and 2:1.

Solution:

(x₁, y₁) = (3, 2)

(x₂, y₂) = (3, 4)

For the ratio 1:2

m : n = 1 : 2

Coordinate of Points = [Tex](\frac{mx_2+nx_1}{m+n},\frac{my_2+ny_1}{m+n})[/Tex]

⇒ Coordinate of Points = (1x₃ + 2x₃/1 + 2, 1x₄ + 2x₂/1 + 2)

⇒ Coordinate of Points = ((3 + 6)/3 , (4 + 4)/3) = (3, 8/3)

Then, for ratio 2:1

Coordinate of Points = [Tex](\frac{mx_2+nx_1}{m+n},\frac{my_2+ny_1}{m+n})[/Tex]

⇒ Coordinate of Points = (2x₃ + 1x₃/2 + 1, 2x₄ + 1x₂/2 + 1)

⇒ Coordinate of Points = ((6 + 3)/3, (8 + 2)/3) = (3, 10/3)

Thus, point of trisections are (3, 8/3) and (3, 10/3).

More Question on Point of Trisections

Question 1. Find the trisection of the points (4,-2) and (3, 1).

Solution:

According to the trisection rule, the line segment with points (4,-2) and (3,1) will be divided into the ratios of 1:2 and 2:1.

Now,

(x₁, y₁) = (4,-2)

(x₂, y₂) = (3,1)

For the ratio 1:2

m : n = 1 : 2

Coordinate of Points = [(mx₂ + nx₁)/(m+n) , (my₂+ny₁)/(m+n)]

⇒ Coordinate of Points = [(1 × 3+2 × 4)/(1+2), (1 × 1+2 × (-2))/(1+2)]

⇒ Coordinate of Points = [(3+8)/3, (1-4)/3] = (11/3, -1)

Then, for the ratio 2:1

m : n = 2 : 1

Coordinate of Points = [(mx₂ + nx₁)/(m+n) , (my₂+ny₁)/(m+n)]

⇒ Coordinate of Points = [(2 × 3 + 1 × 4)/(2 + 1), (2 × 1 + 1 × (-2))/(2+1)]

⇒ Coordinate of Points = (6+4/3, 2-2/3) = (10/3, 0)

Thus, point of trisections are (11/3, -1) and (10/3, 0).

Question 2. Find the trisection of the points (5, -6) and (-7, 5).

Solution:

According to the trisection rule, the line segment with points (5,-6) and (-7,5) will be divided into the ratios of 1:2 and 2:1.

Now,

(x₁, y₁) = (5,-6)

(x₂, y₂) =(-7,5)

For the ratio 1:2

m:n=1:2

Coordinate of Points = [(mx₂ + nx₁)/(m+n) , (my₂+ny₁)/(m+n)]

⇒ Coordinate of Points = (1 × (-7)+2 × 5/1+2, 1 × (5)+2 × (-6)/1+2)

⇒ Coordinate of Points = (-7+10/3, 5-12/3) = (1,-7/2)

For the ratio 2:1

m:n=2:1

Coordinate of Points = [(mx₂ + nx₁)/(m+n) , (my₂+ny₁)/(m+n)]

⇒ Coordinate of Points = (2 × (-7)+1 × 5/2+1, 2 × (5)+1 × (-6)/2+1)

⇒ Coordinate of Points = (-14+5/3, 10-6/3) = (-3,4/3)

Thus, point of trisections are (1,-7/2) and (-3,4/3).

Question 3. Find the trisection of the points (2, 5) and (1, -8).

Solution:

According to the trisection rule, the line segment with points (2,5) and (1,-8) will be divided into the ratios of 1:2 and 2:1.

Now,

(x₁, y₁) = (2,5)

(x₂, y₂) = (1,-8)

For the ratio 1:2

m:n=1:2

Coordinate of Points = [(mx₂ + nx₁)/(m+n) , (my₂+ny₁)/(m+n)]

⇒ Coordinate of Points = (1 × 1+2 × 2/1+2, 1 × (-8)+2 × 5/1+2)

⇒ Coordinate of Points = (1+4/3,-8+10/3)

⇒ Coordinate of Points = (5/3, 2/3)

For the ratio 2:1

m:n=2:1

Coordinate of Points = [(mx₂ + nx₁)/(m+n) , (my₂+ny₁)/(m+n)]

⇒ Coordinate of Points = (2 × 1+1 × 2/2+1, 2 × (-8)+1 × 5/2+1)

⇒ Coordinate of Points = (2+2/3, -16+5/3) = (4/3,-11/3)

Thus, point of trisections are (5/3, 2/3) and (4/3,-11/3).

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Article Tags :

• Mathematics
• Maths MAQ
• School Learning
• Geometry - MAQ