Coordinate Geometry Class 10 Notes (2026-27) — CBSE
Class 10 Maths Chapter 7 notes: the distance formula, the section formula and the midpoint formula, with worked examples and exam question patterns.
Coordinate Geometry — Class 10 Maths Notes
Chapter Snapshot
Coordinate geometry lets you solve geometry using coordinates and algebra. This chapter gives you three tools: the distance formula (how far apart two points are), the section formula (the point dividing a segment in a given ratio), and the midpoint formula. With them you can prove points are collinear, classify triangles and quadrilaterals, and find dividing points.
Board relevance: a scoring chapter with short, formula-driven questions. Expect a distance-formula problem and a section/midpoint problem.
Syllabus note (rationalised): the Area of a Triangle using coordinates has been removed. Focus on the distance and section formulas.
Key Concepts & Definitions
A point in the plane is written as an ordered pair (x, y): x is the abscissa (distance from the y-axis) and y is the ordinate (distance from the x-axis). The axes meet at the origin (0, 0) and split the plane into four quadrants.
Formulas
Distance formula
The distance between P(x₁, y₁) and Q(x₂, y₂):
PQ = √[(x₂ − x₁)² + (y₂ − y₁)²]
- Distance of (x, y) from the origin: √(x² + y²).
- It comes from the Pythagoras theorem applied to the horizontal and vertical gaps.
Section formula (internal division)
The point dividing the segment joining (x₁, y₁) and (x₂, y₂) internally in the ratio m₁ : m₂:
( (m₁x₂ + m₂x₁)/(m₁ + m₂), (m₁y₂ + m₂y₁)/(m₁ + m₂) )
Midpoint formula
The midpoint (ratio 1 : 1):
( (x₁ + x₂)/2, (y₁ + y₂)/2 )
Worked Examples
Example 1 — Distance: Find the distance between A(3, 2) and B(7, 5).
AB = √[(7 − 3)² + (5 − 2)²] = √[16 + 9] = √25 = 5 units.
Example 2 — Type of triangle: Are P(3, 4), Q(−2, 4), R(3, −1) vertices of a right triangle?
PQ = √[(−2−3)² + 0] = 5; PR = √[0 + (−1−4)²] = 5; QR = √[25 + 25] = √50.
Since PQ² + PR² = 25 + 25 = 50 = QR², angle P is 90° → yes, right-angled at P (also isosceles, since PQ = PR).
Example 3 — Section formula: Find the point dividing the segment from A(−1, 3) to B(4, −7) in the ratio 2 : 3.
x = (2·4 + 3·(−1))/(2+3) = (8 − 3)/5 = 1; y = (2·(−7) + 3·3)/5 = (−14 + 9)/5 = −1. Point: (1, −1).
Example 4 — Midpoint: Find the midpoint of A(−2, 6) and B(4, −2).
( (−2 + 4)/2, (6 + (−2))/2 ) = (1, 2). Midpoint: (1, 2).
Example 5 — Collinearity: Show A(1, 2), B(2, 4), C(3, 6) are collinear.
AB = √[1 + 4] = √5; BC = √[1 + 4] = √5; AC = √[4 + 16] = √20 = 2√5.
Since AB + BC = √5 + √5 = 2√5 = AC, the points are collinear.
Example 6 — Find an unknown coordinate: Find the value of y for which the distance between P(2, −3) and Q(10, y) is 10 units.
PQ² = (10 − 2)² + (y + 3)² = 100 → 64 + (y + 3)² = 100 → (y + 3)² = 36 → y + 3 = ±6 → y = 3 or y = −9.
Example 7 — Ratio in which a point divides a segment: In what ratio does the point (−4, 6) divide the segment joining A(−6, 10) and B(3, −8)?
Let the ratio be k : 1. Using the x-coordinate of the section formula: −4 = (3k + (−6))/(k + 1) → −4k − 4 = 3k − 6 → 7k = 2 → k = 2/7. So the ratio is 2 : 7. (Always set the ratio as k : 1 and use one coordinate to solve — the other coordinate can be used to check.)
Example 8 — Point on an axis: Find the ratio in which the x-axis divides the segment joining A(1, −5) and B(−4, 5).
On the x-axis y = 0. Using y = (m₁y₂ + m₂y₁)/(m₁+m₂) = 0 with ratio k:1: (5k − 5)/(k + 1) = 0 → 5k = 5 → k = 1. So the x-axis divides it in the ratio 1 : 1 (at its midpoint).
Important Question Patterns
1. Distance (2 marks): find the distance between two points; distance from the origin; find an unknown coordinate given a distance.
2. Classify a figure (3 marks): use the distance formula to show a triangle is equilateral/isosceles/right-angled, or that a quadrilateral is a parallelogram/rhombus/square (compare sides and diagonals).
3. Section formula (2–3 marks): find the point dividing a segment in a given ratio; find the ratio in which a point (or an axis) divides a segment.
4. Midpoint (2 marks): find a midpoint; use "diagonals of a parallelogram bisect each other" (equal midpoints) to find a missing vertex.
5. Collinearity (2–3 marks): show three points lie on a line using AB + BC = AC.
⚡ Quick Revision
- Distance: PQ = √[(x₂ − x₁)² + (y₂ − y₁)²]; from origin √(x² + y²).
- Section formula (m₁:m₂): ( (m₁x₂ + m₂x₁)/(m₁+m₂), (m₁y₂ + m₂y₁)/(m₁+m₂) ).
- Midpoint: ( (x₁+x₂)/2, (y₁+y₂)/2 ) — section formula with 1:1.
- Right triangle: check whether the sides satisfy a² + b² = c² using the distance formula.
- Collinear points: AB + BC = AC.
- Parallelogram: diagonals bisect each other → their midpoints coincide.
- Removed from syllabus: area of a triangle using coordinates.
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