CBSE Class 10 Maths: Distance & Section Formula Explained

Why distance and section formulas matter

Ever wondered how you can find the exact length between two points on a map, or locate a spot that splits a road in a given ratio? Those tricks are the distance and section formulas – the secret tools behind many CBSE Class 10 geometry questions.

💡 In Simple Words: The distance formula tells you how far apart two points are on a graph, just like measuring the straight‑line distance between two houses. The section formula lets you pinpoint a point that divides a line into a specific ratio, like finding a bench that is 2/5 of the way from one end of a park path.

Distance Formula

What is it?

The distance formula calculates the straight‑line distance between two points (x₁, y₁) and (x₂, y₂) on a coordinate plane. Think of the points as two corners of a rectangle; the distance is the length of the diagonal.

Formula: √[(x₂‑x₁)² + (y₂‑y₁)²]. The square root (√) means “the number that, when multiplied by itself, gives the value inside”.

Step‑by‑step example

Find the distance between A(3, ‑2) and B(‑1, 4).

• Subtract the x‑coordinates: x₂‑x₁ = (‑1)‑3 = ‑4.
• Subtract the y‑coordinates: y₂‑y₁ = 4‑(‑2) = 6.
• Square each difference: (‑4)² = 16, 6² = 36.
• Add them: 16 + 36 = 52.
• Take the square root: √52 ≈ 7.21.

So the distance AB ≈ 7.21 units.

Section Formula

What does it do?

The section formula finds the coordinates of a point that divides a line segment joining two known points in a given ratio. If you imagine a rope tied between two poles, the formula tells you where a knot will be if you pull the rope so that one side is, say, 3 parts and the other side 2 parts.

There are two versions:

• Internal division – the point lies between the two given points.
• External division – the point lies outside the segment, extending the line.

Internal formula (for ratio m:n):

((mx₂ + nx₁)/(m+n), (my₂ + ny₁)/(m+n))

External formula (for ratio m:n):

((mx₂ – nx₁)/(m‑n), (my₂ – ny₁)/(m‑n))

Worked example (internal)

Find the point P that divides the segment joining C(2, 3) and D(8, ‑1) in the ratio 3:2.

• m = 3, n = 2.
• x‑coordinate: (3·8 + 2·2)/(3+2) = (24 + 4)/5 = 28/5 = 5.6.
• y‑coordinate: (3·(‑1) + 2·3)/(3+2) = (‑3 + 6)/5 = 3/5 = 0.6.

Hence P(5.6, 0.6).

Worked example (external)

Find the point Q that divides the line joining E(‑4, 5) and F(6, ‑3) externally in the ratio 1:3.

• m = 1, n = 3.
• x‑coordinate: (1·6 – 3·(‑4))/(1‑3) = (6 + 12)/‑2 = 18/‑2 = ‑9.
• y‑coordinate: (1·(‑3) – 3·5)/(1‑3) = (‑3 – 15)/‑2 = ‑18/‑2 = 9.

Thus Q(‑9, 9).

Quick Comparison

Key Points to Remember

• Always square the differences before adding them in the distance formula.
• Check whether the ratio is internal (use +) or external (use –) before plugging numbers.
• Keep the order of points consistent: m is attached to the second point (x₂, y₂), n to the first (x₁, y₁).
• If the ratio simplifies (e.g., 4:2), reduce it first to avoid big numbers.

📝 Likely Exam Questions

1. Find the distance between (‑2, 5) and (3, ‑1).
   Solution: √[(3‑(‑2))² + (‑1‑5)²] = √[(5)² + (‑6)²] = √(25+36) = √61 ≈ 7.81 units.
2. Point P divides the line joining A(1, 2) and B(7, ‑4) in the ratio 2:3. Find P.
   Solution: P = ((2·7 + 3·1)/(2+3), (2·(‑4) + 3·2)/(2+3)) = ( (14+3)/5 , (‑8+6)/5 ) = (17/5, ‑2/5) = (3.4, ‑0.4).
3. Two points are (4, ‑3) and (‑2, 9). A point Q lies on the line joining them and is 1/4 of the way from the first point. Find Q’s coordinates.
   Solution: Ratio = 1:3 (internal). Q = ((1·(‑2) + 3·4)/(1+3), (1·9 + 3·(‑3))/(1+3)) = ((‑2+12)/4 , (9‑9)/4) = (10/4, 0) = (2.5, 0).
4. Calculate the distance between the points where the line y = x + 2 meets the axes.
   Solution: Intercepts are (0, 2) on y‑axis and (‑2, 0) on x‑axis. Distance = √[(‑2‑0)² + (0‑2)²] = √[4+4] = √8 = 2√2 ≈ 2.83 units.
5. Find the external division point of the segment joining (5, 0) and (‑5, 0) in the ratio 1:2.
   Solution: Using external formula: x = (1·(‑5) – 2·5)/(1‑2) = (‑5‑10)/‑1 = 15, y = (1·0 – 2·0)/(1‑2) = 0. So point is (15, 0).