ICSE Class 10 Maths: Section Formula – Dividing a Line Segment

Why the Section Formula Matters

Ever wondered how to find a point that splits a line exactly where you want? That’s the section formula at work – a handy tool for every coordinate geometry problem.

💡 In Simple Words: The section formula tells you the coordinates of a point that cuts a straight line into a given ratio. Imagine a rope tied between two poles; the formula tells you where to tie a knot so the rope is split 3 : 2, for example.

What is the Section Formula?

The section formula is a rule that gives the coordinates of a point dividing a line segment joining two known points. The two points are usually called A(x₁, y₁) and B(x₂, y₂). The point that splits the segment is called P. If P divides AB in the ratio m:n (read “m to n”), the formula is:

• Internal division (point lies between A and B):
  P(x, y) = \(\frac{mx₂ + nx₁}{m+n}, \frac{my₂ + ny₁}{m+n}\)
• External division (point lies outside the segment):
  P(x, y) = \(\frac{mx₂ - nx₁}{m-n}, \frac{my₂ - ny₁}{m-n}\)

“Internal” means the point is inside the segment, just like a knot on the rope. “External” means the point is outside, like a flag pole placed beyond the rope’s end.

How to Use the Section Formula

Follow these steps whenever a question asks you to find a point that divides a line:

1. Identify the coordinates of the two end points A and B.
2. Read the required ratio m:n from the question.
3. Decide whether the division is internal or external.
4. Plug the numbers into the appropriate formula.
5. Simplify the fractions to get the final coordinates.

Worked Example: Internal Division

Problem: Find the coordinates of point P that divides the line segment joining A(2, 3) and B(8, 9) in the ratio 3 : 2 internally.

Solution:

1. Write down the formula for internal division:
   \(x = \frac{m x_2 + n x_1}{m+n},\; y = \frac{m y_2 + n y_1}{m+n}\)
2. Substitute m = 3, n = 2, \(x_1 = 2, y_1 = 3, x_2 = 8, y_2 = 9\):
   \(x = \frac{3\times8 + 2\times2}{3+2} = \frac{24 + 4}{5} = \frac{28}{5} = 5.6\)
   \(y = \frac{3\times9 + 2\times3}{5} = \frac{27 + 6}{5} = \frac{33}{5} = 6.6\)
3. Thus P = (5.6, 6.6).

Worked Example: External Division

Problem: Find the coordinates of point Q that divides the line joining C(4, 1) and D(10, 7) in the ratio 1 : 3 externally.

Solution:

1. Use the external division formula:
   \(x = \frac{m x_2 - n x_1}{m-n},\; y = \frac{m y_2 - n y_1}{m-n}\)
2. Here m = 1, n = 3, \(x_1 = 4, y_1 = 1, x_2 = 10, y_2 = 7\):
   \(x = \frac{1\times10 - 3\times4}{1-3} = \frac{10 - 12}{-2} = \frac{-2}{-2} = 1\)
   \(y = \frac{1\times7 - 3\times1}{-2} = \frac{7 - 3}{-2} = \frac{4}{-2} = -2\)
3. So Q = (1, -2). Notice Q lies outside the original segment, just as the formula predicts.

Quick Comparison Table

Key Points to Remember

• Always check if the question says “internally” or “externally”.
• Write the ratio as m : n exactly as given – don’t swap them.
• If you get a negative denominator in external division, it’s fine; the sign will cancel out.
• After plugging numbers, simplify step‑by‑step to avoid arithmetic errors.

📝 Likely Exam Questions

1. Question: Find the point that divides the segment joining (1, 2) and (7, 8) in the ratio 4 : 1 internally.
   Answer: Using internal formula, \(x = \frac{4\times7 + 1\times1}{5} = \frac{28+1}{5}=5.8\), \(y = \frac{4\times8 + 1\times2}{5}=\frac{32+2}{5}=6.8\). Point = (5.8, 6.8).
2. Question: A point P divides the line joining (‑3, 5) and (9, ‑1) externally in the ratio 2 : 5. Find P.
   Answer: External formula gives \(x = \frac{2\times9 - 5\times(‑3)}{2‑5}=\frac{18+15}{‑3}=‑11\), \(y = \frac{2\times(‑1) - 5\times5}{‑3}=\frac{‑2‑25}{‑3}=9\). So P = (‑11, 9).
3. Question: If a point Q lies on the line segment joining A(0, 0) and B(10, 10) such that AQ : QB = 3 : 7, find Q’s coordinates.
   Answer: Internal division: \(x = \frac{3\times10 + 7\times0}{10}=3\), \(y = \frac{3\times10 + 7\times0}{10}=3\). Q = (3, 3).