Ever wondered how to find the straight‑line distance between two points on a graph, just like measuring the shortest path on a map?
💡 In Simple Words: The distance formula tells you the length of the straight line joining any two points on a plane. The section formula lets you locate a point that divides a line segment in a given ratio, either inside the segment or outside it.
What is the Distance Formula?
Imagine a flat road that runs from point A to point B. The distance formula is the math version of a ruler that tells you exactly how long that road is, even if the road isn’t horizontal or vertical.
Deriving the formula
Take two points P(x₁, y₁) and Q(x₂, y₂). Draw a right‑angled triangle by dropping a vertical line from Q to the horizontal line through P. The horizontal side has length Δx = x₂ - x₁ and the vertical side has length Δy = y₂ - y₁. By the Pythagorean theorem (a² + b² = c²), the hypotenuse—our straight‑line distance—satisfies:
c² = (Δx)² + (Δy)². Taking the square root gives the distance formula:
Distance = √[(x₂ - x₁)² + (y₂ - y₁)²].
Worked Example 1
Find the distance between A(3, -2) and B(‑1, 4).
- Δx = -1 - 3 = -4
- Δy = 4 - (-2) = 6
Plug into the formula:
Distance = √[(-4)² + 6²] = √[16 + 36] = √52 ≈ 7.21 units.
What is the Section Formula?
The section formula is like a treasure map that tells you exactly where a point lies when you split a line segment in a certain ratio.
Internal vs. external division
If a point P divides the segment joining A(x₁, y₁) and B(x₂, y₂) in the ratio m:n inside the segment, we call it internal division. If P lies on the line extended beyond A or B, that’s external division.
For internal division, the coordinates of P are:
P( (mx₂ + nx₁)/(m+n) , (my₂ + ny₁)/(m+n) ).
For external division, replace the plus signs with minus signs in the numerator:
P( (mx₂ - nx₁)/(m-n) , (my₂ - ny₁)/(m-n) ).
Worked Example 2
Point P divides the line joining C(2, 3) and D(8, ‑1) in the ratio 3:2 internally. Find P’s coordinates.
- 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
So P is at (5.6, 0.6).
Quick Comparison: Distance vs. Section Formula
| Aspect | Distance Formula | Section Formula |
|---|---|---|
| Purpose | Find length of straight line between two points | Find coordinates of a point dividing a segment |
| Key variables | (x₁, y₁), (x₂, y₂) | (x₁, y₁), (x₂, y₂), ratio m:n |
| Typical use in CBSE exams | Geometry questions, distance between cities on a map | Finding mid‑point, locating a point of division, proving collinearity |
Common Mistakes to Avoid
- Forgetting to square the differences before adding them.
- Mixing up (x₂ - x₁) with (x₁ - x₂); the square removes the sign, but it’s good habit to keep order consistent.
- Using the internal formula when the point actually lies outside the segment. Check the problem wording: “extended” or “outside” signals external division.
- Dividing by (m + n) when you need (m - n) for external division.
📝 Likely Exam Questions
- Find the distance between (‑4, 5) and (2, ‑1).
Answer: Δx = 6, Δy = -6 → Distance = √(6² + (-6)²) = √72 = 6√2 units. - Point P divides the line joining (1, 2) and (7, ‑4) in the ratio 1:3 internally. Find P.
Answer: x = (1·7 + 3·1)/(1+3) = (7+3)/4 = 2.5, y = (1·(-4) + 3·2)/4 = (-4+6)/4 = 0.5 → P(2.5, 0.5). - A line segment has endpoints A(‑3, 4) and B(9, ‑2). A point Q divides AB externally in the ratio 2:5. Find Q’s coordinates.
Answer: x = (2·9 – 5·(-3))/(2‑5) = (18 +15)/‑3 = 33/‑3 = -11, y = (2·(-2) – 5·4)/‑3 = (-4 –20)/‑3 = -24/‑3 = 8 → Q(‑11, 8). - Show that the midpoint of the line joining (‑5, 0) and (7, 10) is (1, 5).
Answer: Midpoint = ((‑5+7)/2, (0+10)/2) = (2/2, 10/2) = (1, 5). Hence proved. - Two cities are represented by points (2, 3) and (‑2, ‑1) on a coordinate plane. What is the straight‑line distance between them? (Round off to two decimal places.)
Answer: Δx = -4, Δy = -4 → Distance = √(16+16) = √32 ≈ 5.66 units.