Ever wondered how to find the exact length of a line on a graph without pulling out a ruler?
In simple words, the distance formula tells you the straight‑line length between two points on a plane, while the section formula lets you locate a point that divides a line segment in a given ratio.
What is the Distance Formula?
Imagine you have two cities on a map, with coordinates (x1, y1) and (x2, y2). The distance formula is just the Pythagorean theorem applied to the horizontal and vertical gaps between them.
Formula: Distance = √[(x₂‑x₁)² + (y₂‑y₁)²]
Here, (x₂‑x₁) is the horizontal difference and (y₂‑y₁) is the vertical difference. Squaring each makes them positive, adding them gives the square of the hypotenuse, and the square root brings you back to the actual length.
Worked Example 1
Find the distance between A(3, 4) and B(7, 1).
- Horizontal gap: 7‑3 = 4
- Vertical gap: 1‑4 = –3 (ignore the sign after squaring)
- Apply formula: √[(4)² + (‑3)²] = √[16 + 9] = √25 = 5
So the straight‑line distance AB is 5 units.
Understanding the Section Formula
The section formula helps you find a point that cuts a line segment in a specific ratio, say m:n. Think of a chocolate bar split between two friends: if one gets 3 pieces and the other 2, the cut point divides the bar in the ratio 3:2.
Formula (internal division): If P divides AB in the ratio m:n, then
P(x, y) = ((n·x₁ + m·x₂)/(m + n), (n·y₁ + m·y₂)/(m + n))
When the point lies outside the segment (external division), replace the plus sign in the denominator with a minus sign.
Worked Example 2
Find the coordinates of the point that divides the line joining C(2, ‑1) and D(8, 5) in the ratio 2:3.
- m = 2, n = 3
- x = (3·2 + 2·8)/(2 + 3) = (6 + 16)/5 = 22/5 = 4.4
- y = (3·(‑1) + 2·5)/(2 + 3) = (‑3 + 10)/5 = 7/5 = 1.4
Hence the required point is (4.4, 1.4).
When to Use Which Formula?
| Formula | Typical Use in CBSE Exams |
|---|---|
| Distance formula | Finding the length of a line segment, checking if a triangle is right‑angled, verifying collinearity. |
| Section formula (internal) | Finding mid‑point (special case m=n), locating a point that divides a side in a given ratio, solving coordinate‑geometry word problems. |
| Section formula (external) | Finding points on the extension of a line, solving problems involving external division. |
Quick Checklist
- Identify the two given points and write down their coordinates.
- Decide whether you need a length (distance) or a dividing point (section).
- Plug values carefully – watch the signs!
- Simplify fractions before converting to decimals, unless the question asks for a decimal answer.
Common Mistakes to Avoid
- Mixing up x‑coordinates with y‑coordinates – always keep the pairs together.
- Forgetting to square the differences; a negative gap becomes positive after squaring.
- Using “+” instead of “‑” in the denominator for external division.
- Skipping the square‑root step in the distance formula.
📝 Likely Exam Questions
- Question: Find the distance between the points (‑2, 3) and (4, ‑1).
Answer: √[(4‑(‑2))² + (‑1‑3)²] = √[(6)² + (‑4)²] = √[36 + 16] = √52 = 2√13 units. - Question: The point P divides the line joining (1, 2) and (7, 8) in the ratio 1:2. Find the coordinates of P.
Answer: x = (2·1 + 1·7)/(1 + 2) = (2 + 7)/3 = 3, y = (2·2 + 1·8)/3 = (4 + 8)/3 = 4. So P(3, 4). - Question: Show that the points (2, 3), (5, 7) and (8, 11) are collinear using the distance formula.
Answer: Compute AB = √[(5‑2)² + (7‑3)²] = √[9 + 16] = √25 = 5. BC = √[(8‑5)² + (11‑7)²] = √[9 + 16] = 5. AC = √[(8‑2)² + (11‑3)²] = √[36 + 64] = √100 = 10. Since AB + BC = AC, the three points lie on a straight line. - Question: Find the mid‑point of the line segment joining (‑3, 4) and (9, ‑2).
Answer: Mid‑point is a special case of the section formula with m=n=1: ((‑3+9)/2, (4+‑2)/2) = (3, 1).