Coordinate Geometry — Section Formula

Introduction

The section formula helps you find the coordinates of a point that divides a line segment joining two given points in a given ratio. In ICSE Class 10, you study the internal division formula and its special case — the midpoint formula.

Section Formula (Internal Division)

If point P(x, y) divides the line segment joining A(x₁, y₁) and B(x₂, y₂) internally in the ratio m : n (i.e., AP : PB = m : n), then:

x = (mx₂ + nx₁) / (m + n) y = (my₂ + ny₁) / (m + n)

Alternative Form (Ratio k : 1)

If P divides AB in the ratio k : 1:

x = (kx₂ + x₁) / (k + 1) y = (ky₂ + y₁) / (k + 1)

Midpoint Formula

When the ratio is 1 : 1 (P is the midpoint of AB):

x = (x₁ + x₂) / 2 y = (y₁ + y₂) / 2

Derivation of Section Formula

Let AP : PB = m : n. Draw perpendiculars from A, P, B to the x-axis.

Using similar triangles: (x − x₁) / (x₂ − x) = m / n

Cross-multiplying: n(x − x₁) = m(x₂ − x) nx − nx₁ = mx₂ − mx nx + mx = mx₂ + nx₁ x(m + n) = mx₂ + nx₁ x = (mx₂ + nx₁) / (m + n)

Similarly for y: y = (my₂ + ny₁) / (m + n)

Worked Examples

Example 1: Point Dividing in a Given Ratio

Find the coordinates of the point P that divides the segment joining A(2, 3) and B(7, 8) in the ratio 2 : 3 internally.

Solution: m : n = 2 : 3, A(x₁, y₁) = (2, 3), B(x₂, y₂) = (7, 8)

x = (2 × 7 + 3 × 2) / (2 + 3) = (14 + 6) / 5 = 20 / 5 = 4 y = (2 × 8 + 3 × 3) / (2 + 3) = (16 + 9) / 5 = 25 / 5 = 5

P(4, 5)

Example 2: Midpoint

Find the midpoint of the segment joining P(−3, 4) and Q(5, −2).

Solution: x = (−3 + 5) / 2 = 2 / 2 = 1 y = (4 + (−2)) / 2 = 2 / 2 = 1

Midpoint = (1, 1)

Example 3: Finding the Ratio

In what ratio does the point P(3, 2) divide the segment joining A(1, 0) and B(5, 4)?

Solution: Let AP : PB = m : n.

Using x-coordinate: 3 = (m × 5 + n × 1) / (m + n) 3(m + n) = 5m + n 3m + 3n = 5m + n 2n = 2m m / n = 1 / 1

So the ratio is 1 : 1 (P is the midpoint).

Verify with y-coordinate: y = (1 × 4 + 1 × 0) / (1 + 1) = 4 / 2 = 2 ✓

Example 4: Finding Coordinates of a Vertex

The midpoint of side BC of triangle ABC is D(2, 3). If A is (1, 4), B is (3, 1), find C.

Solution: If D is the midpoint of BC, then: 2 = (3 + x₃) / 2 → 4 = 3 + x₃ → x₃ = 1 3 = (1 + y₃) / 2 → 6 = 1 + y₃ → y₃ = 5

C(1, 5)

Example 5: Points of Trisection

Find the points that trisect the segment joining A(1, −2) and B(4, 7).

Solution: Trisection means dividing into three equal parts. The points divide AB in ratios 1 : 2 and 2 : 1.

First point P (AP : PB = 1 : 2): x = (1 × 4 + 2 × 1) / (1 + 2) = (4 + 2) / 3 = 2 y = (1 × 7 + 2 × (−2)) / (1 + 2) = (7 − 4) / 3 = 1 P(2, 1)

Second point Q (AQ : QB = 2 : 1): x = (2 × 4 + 1 × 1) / (2 + 1) = (8 + 1) / 3 = 3 y = (2 × 7 + 1 × (−2)) / (2 + 1) = (14 − 2) / 3 = 4 Q(3, 4)

Points of trisection: (2, 1) and (3, 4)

Centroid of a Triangle

The centroid G of a triangle with vertices A(x₁, y₁), B(x₂, y₂), C(x₃, y₃) is:

G = [(x₁ + x₂ + x₃) / 3, (y₁ + y₂ + y₃) / 3]

The centroid divides each median in the ratio 2 : 1.

Common Mistakes and Fixes

ICSE Exam Focus

Section formula carries 6–8 marks in ICSE exams. Question types:

• Finding coordinates of a point dividing a segment in a given ratio.
• Finding the ratio in which a point divides a segment.
• Midpoint problems.
• Trisection problems.
• Centroid of a triangle.

Marks Blueprint:

Self-Test Questions

1. Find the coordinates of the point dividing the segment joining A(−3, 5) and B(7, −3) in the ratio 3 : 2.

2. In what ratio does the point (1, 5) divide the line segment joining (−2, 3) and (4, 7)?

3. Find the midpoint of the segment joining (4, −6) and (−2, 8).

4. The points P(1, 2), Q(3, 6), R(7, 4) are the vertices of a triangle. Find the centroid.

5. Find the points of trisection of the segment joining (−2, 1) and (4, 7).

6. If M(2, 3) is the midpoint of AB and A is (5, −1), find B.

In ICSE, remember that the section formula is for internal division only. External division is not in the syllabus.