Most students learn the distance formula and the section formula as two strings of symbols, and they stay as strings until the exam, when one of them is half-remembered and the answer comes out wrong. The good news is that neither formula needs to be memorised from nothing. The distance formula is Pythagoras’ theorem in disguise, and the section formula is the basic proportionality theorem in disguise.
This guide derives each formula in a few lines, works through the standard examples, shows how to use the formulas to identify shapes, and lists the mistakes that cost marks. For the underlying chapter, see the notes on the introduction to coordinate geometry, the distance formula and the section formula. The reasoning for the section formula leans on the basic proportionality theorem and similar triangles.
A Quick Reminder of the Plane
A point on the plane is described by its coordinates (x, y): x is the distance from the y-axis (positive to the right), and y is the distance from the x-axis (positive upwards). The two axes cut the plane into four quadrants. Every problem in this chapter is about the positions of such points and the straight lines between them.
The Distance Formula: Pythagoras in Disguise
Take two points P(x₁, y₁) and Q(x₂, y₂). Draw a horizontal line from P and a vertical line from Q so that they meet at a point R(x₂, y₁). Now PQR is a right-angled triangle with the right angle at R.
- The horizontal side PR has length |x₂ − x₁|.
- The vertical side RQ has length |y₂ − y₁|.
- The slanting side PQ is the distance we want.
By Pythagoras’ theorem, PQ² = PR² + RQ², so
PQ = √[(x₂ − x₁)² + (y₂ − y₁)²]
The absolute value signs disappear because squaring removes the sign. That is why it does not matter which point you call P and which Q.
Example. Find the distance between (3, 4) and (7, 1).
The horizontal difference is 7 − 3 = 4, and the vertical difference is 1 − 4 = −3. Distance = √(16 + 9) = √25 = 5.
The distance from the origin is a special case: for the point (x, y), the distance is √(x² + y²).
The Section Formula: Proportion in Disguise
Suppose a point P divides the segment joining A(x₁, y₁) and B(x₂, y₂) internally in the ratio m : n. That means AP : PB = m : n, with P between A and B.
Drop perpendiculars from A, P and B to the x-axis, meeting it at points with x-coordinates x₁, x and x₂. These three perpendiculars are parallel, so by the basic proportionality idea, they cut every transversal in the same ratio. The horizontal pieces are therefore in the ratio m : n as well:
(x − x₁) : (x₂ − x) = m : n
Solving for x gives n(x − x₁) = m(x₂ − x), so x(m + n) = mx₂ + nx₁. The same reasoning applies to y, which gives the formula:
P = ( (mx₂ + nx₁)/(m + n) , (my₂ + ny₁)/(m + n) )
A way to remember where m and n go: m multiplies the far point’s coordinate (the second point), and n multiplies the near point’s. If the ratio is 1 : 1, the formula becomes the midpoint formula, ((x₁ + x₂)/2, (y₁ + y₂)/2).
Example. Find the point that divides the segment joining (−1, 7) and (4, −3) in the ratio 2 : 3.
x = (2 × 4 + 3 × (−1))/5 = (8 − 3)/5 = 1. y = (2 × (−3) + 3 × 7)/5 = (−6 + 21)/5 = 3. The point is (1, 3).
Finding the ratio. In what ratio does the y-axis divide the segment joining (−4, −6) and (10, 8)?
On the y-axis, x = 0. Let the ratio be k : 1. Then 0 = (10k + 1 × (−4))/(k + 1), so 10k = 4 and k = 2/5. The ratio is 2 : 5. The y-coordinate of the point is (8 × 2/5 + (−6))/(7/5) = −2, so the point is (0, −2).
Using the Formulas to Identify Shapes
The formulas become powerful when they are used to decide what shape some points make.
| To show that… | Use | What to check |
| Three points are collinear | Distance formula | The two shorter distances add up to the longest |
| A triangle is isosceles | Distance formula | Two sides are equal |
| A triangle is right-angled | Distance formula and Pythagoras | The square of the longest side equals the sum of the squares of the other two |
| A quadrilateral is a parallelogram | Midpoint formula | The diagonals have the same midpoint |
| A quadrilateral is a rhombus | Distance formula | All four sides are equal |
| A quadrilateral is a square | Distance formula | All four sides are equal, and the two diagonals are equal |
Example. Show that A(1, 7), B(4, 2), C(−1, −1) and D(−4, 4) are the vertices of a square.
AB² = 9 + 25 = 34, BC² = 25 + 9 = 34, CD² = 9 + 25 = 34 and DA² = 25 + 9 = 34. All four sides are equal. The diagonals are AC² = 4 + 64 = 68 and BD² = 64 + 4 = 68, which are also equal. A quadrilateral with all sides equal and equal diagonals is a square. ✓
A good habit is to work with the squares of the distances, and take the square root only when you need the actual length. It saves effort and avoids errors with surds.
Where This Is Used
Coordinates are how maps, screens and navigation systems describe positions. On a city map drawn on a grid, the distance formula gives the straight-line distance between two places, though roads rarely run in straight lines. Computer games and phone screens place every object by its coordinates and use the same formula to decide whether two objects have collided. The idea is the same at every scale: a position is a pair of numbers, and a distance is what the Pythagorean theorem makes of them.
Slips That Cost Marks
- Subtracting x from y, mixing the order of coordinates inside the brackets.
- Forgetting the square root at the end, and giving the squared distance.
- Putting m with the wrong point in the section formula, which gives the ratio n : m by mistake.
- Using the section formula for an external division without noticing that the ratio is negative.
- Not simplifying surds, such as leaving √68 instead of 2√17.
- Writing the answer for a point as two separate numbers instead of an ordered pair.
Coordinate Geometry in Your Syllabus
Coordinate Geometry is a chapter of the Class 10 Maths book in both the CBSE and the Karnataka SSLC textbooks. Published deleted-syllabus lists show that CBSE has trimmed the area of a triangle in this chapter, while the distance and section formulas remain, so check your own board’s syllabus for that part. Practise with the chapter MCQs, the important questions and the chapter notes. In Class 11, the topic returns in straight lines and conic sections, as in these notes on the coordinates of a point, which matter for JEE Main and KCET.
Frequently Asked Questions
Q1. What is the distance formula?
The distance between (x₁, y₁) and (x₂, y₂) is √[(x₂ − x₁)² + (y₂ − y₁)²].
Q2. Where does the distance formula come from?
From Pythagoras’ theorem, applied to a right triangle whose legs are the horizontal and vertical differences between the two points.
Q3. What is the section formula?
The point dividing the segment joining (x₁, y₁) and (x₂, y₂) internally in the ratio m : n is ((mx₂ + nx₁)/(m + n), (my₂ + ny₁)/(m + n)).
Q4. How do I find the midpoint of a segment?
Use the section formula with m = n = 1, which gives ((x₁ + x₂)/2, (y₁ + y₂)/2).
Q5. How can I show that a quadrilateral is a parallelogram?
Show that the midpoints of its two diagonals are the same point.
A Formula You Can Rebuild
When a formula comes from an idea, you can rebuild it from the idea if you forget it. Pythagoras gives the distance, proportion gives the section point, and the rest is careful arithmetic with signs.
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