Understanding the Section Formula

An interactive guide to coordinate geometry.

A Real-World Scenario

Hello! Welcome to this interactive session. Get ready to explore how we can find specific points on a line segment.

Let's start with a simple real-world scenario. Imagine a straight road connecting two villages, Village A and Village B. A bridge is built on this road that divides the total distance in a specific ratio.

Why Do We Need a Formula for This?

Let's say the road is 100 km long. A bridge, point C, divides the road in the ratio of 3:7. It's easy to see that Village B is farther away because the larger number (7) corresponds to the longer distance. We can even calculate the exact distances:

Distance from A to C = \( \frac{3}{3+7} \times 100 = 30 \) km

Distance from B to C = \( \frac{7}{3+7} \times 100 = 70 \) km

This is straightforward on a simple line. But what if we place this road on a map with a coordinate system (with x and y axes)?

Adding Coordinates

Now, let's place our villages on a graph. Village A is at \( (x_1, y_1) \) and Village B is at \( (x_2, y_2) \). We want to find the exact coordinates of the bridge, let's call it point C (x, y), which divides the line segment AB in the ratio m:n.

This is where the section formula becomes incredibly useful. The formula looks like this:

$$ C(x,y) = \left( \frac{mx_2 + nx_1}{m+n}, \frac{my_2 + ny_1}{m+n} \right) $$

Let's Try An Example

Suppose Village A is at (2, 3) and Village B is at (9, 10). The bridge C divides the road in the ratio 3:4. Using the values: \(x_1=2, y_1=3, x_2=9, y_2=10, m=3, n=4\), let's find the coordinates of the bridge C.

x-coordinate: \( x = \frac{(3 \times 9) + (4 \times 2)}{3+4} = \frac{27+8}{7} = \frac{35}{7} = 5 \)

y-coordinate: \( y = \frac{(3 \times 10) + (4 \times 3)}{3+4} = \frac{30+12}{7} = \frac{42}{7} = 6 \)

So, the exact coordinates of the bridge are (5, 6).

So, Why Do We Use the Section Formula?

We use it to find the exact coordinates of a point that divides a line segment into a specific ratio. It's a fundamental tool in coordinate geometry used for:

• Navigation and GPS: Calculating intermediate points on a route.
• Computer Graphics: Determining points for drawing lines and shapes.
• Physics: Finding the center of mass of systems.

What is the Section Formula?

In coordinate geometry, the section formula is used to find the coordinates of a point that divides a line segment into two parts in a given ratio. Imagine you have a line segment between two points, A and B. The section formula helps you find the exact location of a point P on that line segment, such that the ratio of the length of AP to the length of PB is a specific value, say \( m_1 : m_2 \).

This is incredibly useful in many fields, including physics (for finding the center of mass), computer graphics (for calculating paths and positions), and of course, for solving a wide range of geometry problems.

Interactive Visualizer

Resulting Point P(x, y)

Control Panel

Point A (\(x_1, y_1\))

x₁: 2

y₁: 3

Point B (\(x_2, y_2\))

x₂: 8

y₂: 7

Ratio (\(m_1\) : \(m_2\))

m₁: 1

m₂: 2

Interactive Formula Breakdown

See how the values from the sliders plug into the formula. The colors match the points and ratio values from the interactive graph above.

P(x,y) = (

(1 x 8) + (2 x 2) 1 + 2

,

(1 x 7) + (2 x 3) 1 + 2

)

P = ( , )

Click here for a step-by-step derivation

The derivation uses similar triangles. The diagram on the right is generated live from the values in the control panel. Try moving the sliders!

1. Draw perpendiculars from A, P, and B to the x-axis (AR, PS, BT). Draw lines from A and P parallel to the x-axis, creating points Q and C.

2. This creates two right-angled triangles, \(\triangle AQP\) and \(\triangle PCB\). They are similar (\(\triangle AQP \sim \triangle PCB\)) because they share the same angles (AA similarity).

3. Since the triangles are similar, the ratio of their corresponding sides is equal:

$$ \frac{AP}{PB} = \frac{AQ}{PC} = \frac{QP}{CB} $$

4. We know that \( \frac{AP}{PB} = \frac{m_1}{m_2} \). Now let's express the lengths of the other sides in terms of coordinates:

• \( AQ = RS = OS - OR = x - x_1 \)
• \( PC = ST = OT - OS = x_2 - x \)
• \( QP = PS - QS = y - y_1 \)
• \( CB = BT - CT = y_2 - y \)

5. Substitute these values back into the ratio equation:

$$ \frac{m_1}{m_2} = \frac{x - x_1}{x_2 - x} \quad \text{and} \quad \frac{m_1}{m_2} = \frac{y - y_1}{y_2 - y} $$

6. Now, solve for x and y separately. Let's start with the x-coordinate:

$$ m_1(x_2 - x) = m_2(x - x_1) \implies m_1x_2 - m_1x = m_2x - m_2x_1 $$ $$ m_1x_2 + m_2x_1 = x(m_1 + m_2) \implies x = \frac{m_1x_2 + m_2x_1}{m_1 + m_2} $$

7. Similarly, solving for the y-coordinate gives:

$$ y = \frac{m_1y_2 + m_2y_1}{m_1 + m_2} $$

And that's how we arrive at the section formula!

$$ P(x, y) = \left( \frac{m_1x_2 + m_2x_1}{m_1 + m_2}, \frac{m_1y_2 + m_2y_1}{m_1 + m_2} \right) $$

Special Case: The Midpoint Formula

A very common use case is finding the midpoint of a line segment. The midpoint divides the segment into two equal halves, meaning the ratio is \(1:1\). So, we have \(m_1 = 1\) and \(m_2 = 1\).

If we substitute these values into the section formula:

$$ P(x, y) = \left( \frac{1 \cdot x_2 + 1 \cdot x_1}{1 + 1}, \frac{1 \cdot y_2 + 1 \cdot y_1}{1 + 1} \right) = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) $$

This simplified formula is known as the Midpoint Formula.

Solved Examples

Example 1:

Find the coordinates of the point which divides the line segment joining the points A(4, -3) and B(8, 5) in the ratio 3:1 internally.

Show Solution

Given:
A(\(x_1, y_1\)) = (4, -3)
B(\(x_2, y_2\)) = (8, 5)
Ratio \(m_1 : m_2 = 3 : 1\)

Using the section formula:

$$ x = \frac{m_1x_2 + m_2x_1}{m_1 + m_2} = \frac{3(8) + 1(4)}{3 + 1} = \frac{24 + 4}{4} = \frac{28}{4} = 7 $$

$$ y = \frac{m_1y_2 + m_2y_1}{m_1 + m_2} = \frac{3(5) + 1(-3)}{3 + 1} = \frac{15 - 3}{4} = \frac{12}{4} = 3 $$

Answer: The coordinates of the point are (7, 3).

Example 2:

In what ratio does the point P(2, -5) divide the line segment joining the points A(-3, 5) and B(4, -9)?

Show Solution

Given:
A(\(x_1, y_1\)) = (-3, 5)
B(\(x_2, y_2\)) = (4, -9)
P(x, y) = (2, -5)

Let the ratio be \(k : 1\). So, \(m_1 = k\) and \(m_2 = 1\).

Using the section formula for the x-coordinate:

$$ x = \frac{kx_2 + 1x_1}{k + 1} \implies 2 = \frac{k(4) + 1(-3)}{k + 1} $$

$$ 2(k + 1) = 4k - 3 \implies 2k + 2 = 4k - 3 \implies 5 = 2k \implies k = \frac{5}{2} $$

The ratio is \(k:1\), which is \(\frac{5}{2} : 1\), or \(5:2\).

Answer: The ratio is 5:2.

Example 3 (Midpoint):

Find the midpoint of the line segment joining the points A(-5, 4) and B(3, -2).

Show Solution

Given:
A(\(x_1, y_1\)) = (-5, 4)
B(\(x_2, y_2\)) = (3, -2)

Using the midpoint formula:

$$ x = \frac{x_1 + x_2}{2} = \frac{-5 + 3}{2} = \frac{-2}{2} = -1 $$

$$ y = \frac{y_1 + y_2}{2} = \frac{4 + (-2)}{2} = \frac{2}{2} = 1 $$

Answer: The midpoint of the line segment is (-1, 1).