Why distance and section formulas matter
Ever wondered how to find the exact length of a road on a map or the spot where a treasure is hidden between two landmarks? That's what the distance and section formulas do in coordinate geometry.
💡 In Simple Words: The distance formula tells you how far two points are on a graph, just like measuring a straight line with a ruler. The section formula tells you the exact coordinates of a point that splits a line segment into a given ratio, like cutting a candy bar into two unequal pieces.
What is the distance formula?
In a rectangular (Cartesian) plane, each point has an (x, y) pair. To find the straight‑line distance between points P(x₁, y₁) and Q(x₂, y₂), imagine a right‑angled triangle with the two points as corners. The horizontal side is the difference in x‑values, the vertical side is the difference in y‑values. By the Pythagorean theorem (a² + b² = c²), the hypotenuse c is the distance.
The formula becomes:
Distance = √[(x₂‑x₁)² + (y₂‑y₁)²]
Worked example: distance between (3, ‑2) and (‑1, 4)
- Find the differences: x₂‑x₁ = (‑1)‑3 = ‑4, y₂‑y₁ = 4‑(‑2) = 6.
- Square them: (‑4)² = 16, 6² = 36.
- Add: 16 + 36 = 52.
- Take the square root: √52 ≈ 7.21 units.
So the two points are about 7.2 units apart.
What is the section formula?
The section formula helps you locate a point that divides a line segment joining A(x₁, y₁) and B(x₂, y₂) in a given ratio m:n. Think of a chocolate bar that you break so that one piece is m parts and the other is n parts. The point of the break has coordinates that are weighted averages of the ends.
If the point P is internally (inside the segment) dividing AB in the ratio m:n, the coordinates are:
P(x, y) = ( (n·x₁ + m·x₂)/(m + n) , (n·y₁ + m·y₂)/(m + n) )
For an external division (outside the segment), the signs of m or n change, but CBSE Class 10 only asks for the internal case.
Worked example: point dividing (2, 3) and (8, ‑1) in the ratio 3:2
- Plug into the formula: x = (2·2 + 3·8)/(3 + 2) = (4 + 24)/5 = 28/5 = 5.6
- y = (2·3 + 3·(‑1))/(3 + 2) = (6 ‑ 3)/5 = 3/5 = 0.6
Hence the required point is (5.6, 0.6).
Quick comparison table
| Formula | When to use |
|---|---|
| √[(x₂‑x₁)² + (y₂‑y₁)²] | Finding straight‑line distance between two points |
| ( (n·x₁ + m·x₂)/(m + n) , (n·y₁ + m·y₂)/(m + n) ) | Finding a point that divides a segment in the ratio m:n |
Common mistakes to avoid
- Mixing up x‑difference and y‑difference – always subtract in the same order (x₂‑x₁, y₂‑y₁).
- Forgetting to square the differences before adding.
- Using m and n in the wrong places in the section formula – the numerator pairs the opposite end’s coordinate with the opposite ratio.
- Skipping the square‑root step in the distance formula.
📝 Likely Exam Questions
- Find the distance between points (‑3, 5) and (4, ‑2).
Answer: √[(4‑(‑3))² + (‑2‑5)²] = √[7² + (‑7)²] = √[49 + 49] = √98 ≈ 9.9 units. - Point P divides the line joining A(1, 2) and B(7, ‑4) in the ratio 1:3. Find the coordinates of P.
Answer: x = (3·1 + 1·7)/4 = (3 + 7)/4 = 2.5, y = (3·2 + 1·(‑4))/4 = (6‑4)/4 = 0.5 → P(2.5, 0.5). - Two points are (2, 3) and (‑2, ‑1). A point Q is 2/5 of the way from the first point to the second. Find Q’s coordinates.
Answer: Ratio m:n = 2:3 (since Q is 2 parts from first point). x = (3·2 + 2·(‑2))/5 = (6‑4)/5 = 0.4, y = (3·3 + 2·(‑1))/5 = (9‑2)/5 = 1.4 → Q(0.4, 1.4). - Show that the distance between (x₁, y₁) and (x₂, y₂) is the same as the distance between (x₂, y₂) and (x₁, y₁).
Answer: Plugging into the formula gives √[(x₂‑x₁)² + (y₂‑y₁)²] which is identical after swapping the order because squaring removes the sign.