Trisection Points of a Line Segment
An interactive guide to understanding and calculating the points that divide a line segment into three equal parts.
What are Trisection Points?
In coordinate geometry, the points of trisection are two specific points on a line segment that divide it into three equal parts.
Imagine you have a line segment AB. The two trisection points, let's call them P and Q, would be located such that the length of AP is equal to the length of PQ, which is also equal to the length of QB.
To find these points, we use the Section Formula. Point P divides the line segment AB in the ratio 1:2, and point Q divides it in the ratio 2:1.
The Formula Explained
The trisection formulas are a specific application of the general Section Formula. The section formula is used to find the coordinates of a point that divides a line segment in a given ratio, $m_1:m_2$.
The General Section Formula
For a line segment with endpoints A($x₁, y₁$) and B($x₂, y₂$), the point P that divides it in the ratio $m_1:m_2$ is:
$$ P(x, y) = \left( \frac{m_1 x_2 + m_2 x_1}{m_1+m_2} , \frac{m_1 y_2 + m_2 y_1}{m_1+m_2} \right) $$
To find the trisection points, we apply this formula with two different ratios:
Point P (ratio $m_1:m_2$ = 1:2)
$$ P(x, y) = \left( \frac{1 \cdot x_2 + 2 \cdot x_1}{1+2} , \frac{1 \cdot y_2 + 2 \cdot y_1}{1+2} \right) $$
$$ P(x, y) = \left( \frac{x_2 + 2x_1}{3} , \frac{y_2 + 2y_1}{3} \right) $$
Point Q (ratio $m_1:m_2$ = 2:1)
$$ Q(x, y) = \left( \frac{2 \cdot x_2 + 1 \cdot x_1}{2+1} , \frac{2 \cdot y_2 + 1 \cdot y_1}{2+1} \right) $$
$$ Q(x, y) = \left( \frac{2x_2 + x_1}{3} , \frac{2y_2 + y_1}{3} \right) $$
Interactive Calculator
Test Your Knowledge
Question 1
Find the coordinates of the points of trisection of the line segment joining the points (4, -1) and (-2, -3).
Let A = (4, -1) and B = (-2, -3).
For point P (ratio 1:2):
$$ x = \frac{-2 + 2(4)}{3} = \frac{-2 + 8}{3} = \frac{6}{3} = 2 $$
$$ y = \frac{-3 + 2(-1)}{3} = \frac{-3 - 2}{3} = -\frac{5}{3} $$
So, $P = (2, -\frac{5}{3})$
For point Q (ratio 2:1):
$$ x = \frac{2(-2) + 4}{3} = \frac{-4 + 4}{3} = \frac{0}{3} = 0 $$
$$ y = \frac{2(-3) + (-1)}{3} = \frac{-6 - 1}{3} = -\frac{7}{3} $$
So, $Q = (0, -\frac{7}{3})$
The points are $(2, -\frac{5}{3})$ and $(0, -\frac{7}{3})$.
Question 2
Find the points of trisection for the line segment with endpoints P(0, 5) and Q(9, -4).
Let A = (0, 5) and B = (9, -4).
For the first point (ratio 1:2):
$$ x = \frac{9 + 2(0)}{3} = \frac{9}{3} = 3 $$
$$ y = \frac{-4 + 2(5)}{3} = \frac{-4 + 10}{3} = \frac{6}{3} = 2 $$
So, the first point is $(3, 2)$
For the second point (ratio 2:1):
$$ x = \frac{2(9) + 0}{3} = \frac{18}{3} = 6 $$
$$ y = \frac{2(-4) + 5}{3} = \frac{-8 + 5}{3} = \frac{-3}{3} = -1 $$
So, the second point is $(6, -1)$
The points are $(3, 2)$ and $(6, -1)$.
Question 3 (Challenge)
The point P(5, -2) is one of the points of trisection of the line segment joining the points A(7, -4) and B(1, 2). Is P closer to A or B? Justify your answer.
We need to check which ratio (1:2 or 2:1) gives the point P(5, -2).
Case 1: P divides AB in the ratio 1:2.
$$ x = \frac{1(1) + 2(7)}{3} = \frac{1 + 14}{3} = \frac{15}{3} = 5 $$
$$ y = \frac{1(2) + 2(-4)}{3} = \frac{2 - 8}{3} = \frac{-6}{3} = -2 $$
This gives the point $(5, -2)$, which is our point P.
Since P divides the segment AB in the ratio 1:2, it means the distance AP is one-third of the total distance AB, while the distance PB is two-thirds.
Therefore, P(5, -2) is closer to A.