The Section Formula and Points of Trisection for CBSE Class 10

While the distance formula allows us to measure the total length between two points, geometry frequently requires us to locate intermediate points along a line segment. For instance, what are the coordinates of a milestone located one-third of the way between two cities, or where does a communication tower divide the transmission line between two relay stations?

In CBSE Class 10 Mathematics, the Section Formula provides the exact algebraic coordinates of a point that divides a line segment into any given internal ratio $m_1 : m_2$. Mastering the section formula—along with the clever $k : 1$ method—is essential for tackling high-weightage questions in board examinations.

What You Will Learn

• Statement and algebraic structure of the Section Formula for internal division
• The criss-cross multiplication memory technique
• The $k : 1$ method for finding unknown division ratios
• Finding the ratio in which the $x$-axis or $y$-axis divides a line segment
• Determining the points of trisection of a line segment
• Step-by-step solved CBSE board examination problems and common traps

1. The Section Formula (Internal Division)

Let $A(x_1, y_1)$ and $B(x_2, y_2)$ be two given points in the Cartesian coordinate plane. Let $P(x, y)$ be a point on the line segment $AB$ that divides it internally in the ratio $m_1 : m_2$, such that: $\frac{AP}{PB} = \frac{m_1}{m_2}$

             A(x1, y1) --------- P(x, y) ----------------- B(x2, y2)
                         m1                 m2

The Section Formula

The coordinates of the point $P(x, y)$ dividing the line segment joining $A(x_1, y_1)$ and $B(x_2, y_2)$ internally in the ratio $m_1 : m_2$ are given by: $P(x, y) = \left( \frac{m_1 x_2 + m_2 x_1}{m_1 + m_2}, \quad \frac{m_1 y_2 + m_2 y_1}{m_1 + m_2} \right)$

The Criss-Cross Memory Rule:

Notice the cross-multiplication pattern:

• The ratio part on the left ($m_1$) multiplies the coordinate on the right ($x_2$ and $y_2$).
• The ratio part on the right ($m_2$) multiplies the coordinate on the left ($x_1$ and $y_1$).
• The denominator is always the sum of the ratio parts: $m_1 + m_2$.

$\mathbf{x = \frac{m_1 x_2 + m_2 x_1}{m_1 + m_2}}, \quad \mathbf{y = \frac{m_1 y_2 + m_2 y_1}{m_1 + m_2}}$

2. The $k : 1$ Ratio Method

When a board question asks you to find the ratio in which a given point $P$ divides the line segment $AB$:

• Working with two unknown variables $m_1$ and $m_2$ creates unnecessary complexity.
• Since $\frac{m_1}{m_2} = \frac{m_1 / m_2}{1} = \frac{k}{1}$, we assume the ratio is $k : 1$.
• This reduces the problem to solving for a single unknown variable $k$!

The Formula with Ratio $k : 1$:

$P(x, y) = \left( \frac{k x_2 + x_1}{k + 1}, \quad \frac{k y_2 + y_1}{k + 1} \right)$

Once $k$ is found, the required ratio is $k : 1$. For example, if $k = \frac{2}{7}$, the ratio is $2 : 7$.

3. Division by the Coordinate Axes (CBSE Favorite)

1. Division by the $x$-axis:
   • Any point on the $x$-axis has its $y$-coordinate equal to zero: $P(x, 0)$.
   • Equate the $y$-expression of the section formula to zero: $\frac{m_1 y_2 + m_2 y_1}{m_1 + m_2} = 0 \implies m_1 y_2 + m_2 y_1 = 0$
   • This immediately yields the ratio!
2. Division by the $y$-axis:
   • Any point on the $y$-axis has its $x$-coordinate equal to zero: $P(0, y)$.
   • Equate the $x$-expression to zero: $\frac{m_1 x_2 + m_2 x_1}{m_1 + m_2} = 0 \implies m_1 x_2 + m_2 x_1 = 0$

4. Points of Trisection of a Line Segment

Points of trisection are two points that divide a line segment into three equal parts.

             A(x1, y1) ----- P ----- Q ----- B(x2, y2)
                 1           1       1

If $P$ and $Q$ trisect the segment $AB$:

• Point $P$ divides $AB$ internally in the ratio $1 : 2$ ($AP = 1$ part, $PB = 2$ parts).
• Point $Q$ divides $AB$ internally in the ratio $2 : 1$ ($AQ = 2$ parts, $QB = 1$ part).
• Alternatively, once point $P$ is found, point $Q$ is simply the mid-point of segment $PB$!

5. Solved CBSE Board Examination Problems

Solved Example 1: Direct Section Coordinates

Problem: Find the coordinates of the point which divides the join of $(-1, 7)$ and $(4, -3)$ in the ratio $2 : 3$.

Solution:

1. Here $x_1 = -1, y_1 = 7$; $x_2 = 4, y_2 = -3$; and $m_1 = 2, m_2 = 3$.
2. Apply the Section Formula: $x = \frac{m_1 x_2 + m_2 x_1}{m_1 + m_2} = \frac{2(4) + 3(-1)}{2 + 3} = \frac{8 - 3}{5} = \frac{5}{5} = 1$ $y = \frac{m_1 y_2 + m_2 y_1}{m_1 + m_2} = \frac{2(-3) + 3(7)}{2 + 3} = \frac{-6 + 21}{5} = \frac{15}{5} = 3$
3. Therefore, <u>the coordinates of the required point are $(1, 3)$</u>.

Solved Example 2: Points of Trisection (NCERT Classic)

Problem: Find the coordinates of the points of trisection of the line segment joining $A(2, -2)$ and $B(-7, 4)$.

Solution: Let $P$ and $Q$ be the points of trisection of $AB$.

1. Finding Point $P$ (Ratio $1 : 2$): $x_P = \frac{1(-7) + 2(2)}{1 + 2} = \frac{-7 + 4}{3} = \frac{-3}{3} = -1$ $y_P = \frac{1(4) + 2(-2)}{1 + 2} = \frac{4 - 4}{3} = \frac{0}{3} = 0$ Thus, $P(-1, 0)$.
2. Finding Point $Q$ (Ratio $2 : 1$): $x_Q = \frac{2(-7) + 1(2)}{2 + 1} = \frac{-14 + 2}{3} = \frac{-12}{3} = -4$ $y_Q = \frac{2(4) + 1(-2)}{2 + 1} = \frac{8 - 2}{3} = \frac{6}{3} = 2$ Thus, $Q(-4, 2)$.
3. Therefore, <u>the points of trisection are $(-1, 0)$ and $(-4, 2)$</u>.

Solved Example 3: Ratio Divided by the x-axis (CBSE PYQ)

Problem: Find the ratio in which the line segment joining $A(1, -5)$ and $B(-4, 5)$ is divided by the $x$-axis. Also find the coordinates of the point of division.

Solution:

1. Let the $x$-axis divide $AB$ at point $P(x, 0)$ in the ratio $k : 1$.
2. By the section formula for the $y$-coordinate: $y = \frac{k(5) + 1(-5)}{k + 1}$
3. Since $P$ lies on the $x$-axis, its $y$-coordinate is $0$: $\frac{5k - 5}{k + 1} = 0 \implies 5k - 5 = 0 \implies 5k = 5 \implies k = 1$ Therefore, <u>the $x$-axis divides $AB$ in the ratio $1 : 1$</u> (i.e., $P$ is the mid-point of $AB$).
4. Find the $x$-coordinate of $P$: $x = \frac{1(-4) + 1(1)}{1 + 1} = \frac{-4 + 1}{2} = -\frac{3}{2}$
5. Therefore, <u>the coordinates of the point of division are $\left(-\frac{3}{2}, 0\right)$</u>.

6. Summary and Examination Tips

Exam Tip: In questions asking for points of trisection, always state that "trisection means dividing into three equal parts, which creates two points dividing the segment in ratios $1:2$ and $2:1$".

Common Mistake: Swapping ratio parts! Remember that $m_1$ multiplies $(x_2, y_2)$ and $m_2$ multiplies $(x_1, y_1)$. Multiplying $m_1$ by $(x_1, y_1)$ inverts the division!