Section Formula Class 10 Maths: Examples & Questions

The section formula is used to find the coordinates of a point that divides a line segment joining two given points in a specific ratio. Instead of relying on diagrams, this formula allows you to directly calculate the required point using the coordinates of the given endpoints.

Table of Contents

• What is the Section Formula
• Solved Examples on Section Formula

What is the Section Formula

If a point P(x, y) divides the line segment joining A(x₁, y₁) and B(x₂, y₂) in the ratio m : n, the section formula gives the coordinates of P.

P(x, y) = ((m x₂ + n x₁) / (m + n), (m y₂ + n y₁) / (m + n))

When point P lies between A and B, it is called internal division.

Proof:

Consider two points A(x₁, y₁) and B(x₂, y₂). Assume P(x, y) divides AB internally in the ratio m₁ : m₂.

PA / PB = m₁ / m₂

From similarity of triangles: PA / PB = AQ / PC = PQ / BC

• AQ = x − x₁
• PC = x₂ − x
• PQ = y − y₁
• BC = y₂ − y

So, m₁ / m₂ = (x − x₁) / (x₂ − x) = (y − y₁) / (y₂ − y)

• x = (m₁x₂ + m₂x₁) / (m₁ + m₂)
• y = (m₁y₂ + m₂y₁) / (m₁ + m₂)

((m₁x₂ + m₂x₁) / (m₁ + m₂), (m₁y₂ + m₂y₁) / (m₁ + m₂))

NOTE: The midpoint divides the line segment in the ratio 1 : 1.

Midpoint = ((x₁ + x₂) / 2, (y₁ + y₂) / 2)

Solved Examples on Section Formula

Example 1: Point P divides A(4, 6) and B(−5, −4) internally in ratio 3 : 2.

Solution:

x = (3 × −5 + 2 × 4) / (3 + 2) = (−15 + 8) / 5 = −7/5

y = (3 × −4 + 2 × 6) / (3 + 2) = (−12 + 12) / 5 = 0

∴ P = (−7/5, 0)

Example 2: Point (−4, 6) divides A(−6, 10) and B(3, −8). Find the ratio.

Solution:

Let ratio = k : 1

−4 = (3k − 6) / (k + 1)

−4(k + 1) = 3k − 6

−4k − 4 = 3k − 6

2 = 7k ⇒ k = 2/7

Ratio = 2 : 7

Centroid: G = ((x₁ + x₂ + x₃) / 3, (y₁ + y₂ + y₃) / 3)

Gx = (−1 + 2 + 8) / 3 = 3

Gy = (−3 + 1 − 4) / 3 = −2

Centroid G = (3, −2)