By the end of this chapter you'll be able to…

  • 1Apply distance formula
  • 2Use section formula for internal division
  • 3Calculate area of triangle from coordinates
  • 4Check collinearity
  • 5Find midpoint and centroid
💡
Why this chapter matters
Foundation for Class 11 conic sections, Class 12 calculus, all of physics and engineering.

Before you start — revise these

A 5-minute refresher here will save you 30 minutes of confusion below.

Coordinate Geometry — Class 10 Mathematics

"Geometry and algebra unite — a point becomes a pair of numbers, a line becomes an equation."

1. About the Chapter

Coordinate Geometry (or Cartesian Geometry) describes geometric shapes using algebra. Founded by René Descartes (1637) and earlier influenced by Indian mathematicians.

Key Idea

Every point in the plane is identified by (x, y) — ordered pair of coordinates.

What This Chapter Covers

  • Recap of Cartesian plane
  • Distance Formula
  • Section Formula (internal division)
  • Area of Triangle from coordinates
  • Applications

2. The Cartesian Plane (Quick Review)

Axes

  • x-axis: horizontal
  • y-axis: vertical
  • They meet at ORIGIN (0, 0)

Quadrants

  • Q1: top-right, (+x, +y)
  • Q2: top-left, (−x, +y)
  • Q3: bottom-left, (−x, −y)
  • Q4: bottom-right, (+x, −y)

Plot a Point

P(3, 5): go 3 units right, 5 units up. Q(−2, −4): go 2 units left, 4 units down.


3. Distance Formula

Statement

Distance between P(x₁, y₁) and Q(x₂, y₂):

PQ = √((x₂−x₁)² + (y₂−y₁)²)

Derivation

Form right triangle: horizontal leg = (x₂−x₁), vertical leg = (y₂−y₁). Apply Pythagoras.

Examples

Example 1: Find distance between A(2, 3) and B(5, 7).

  • PQ = √((5−2)² + (7−3)²) = √(9 + 16) = √25 = 5 units

Example 2: Find distance from origin (0,0) to P(−3, 4).

  • = √(9 + 16) = √25 = 5 units

Example 3: Verify (0,0), (3,4), (−3,4) form an isosceles triangle.

  • d₁ from (0,0) to (3,4): 5
  • d₂ from (0,0) to (−3,4): 5
  • d₃ from (3,4) to (−3,4): 6
  • Two sides equal (5 = 5), so isosceles ✓

4. Section Formula (Internal Division)

Statement

If P(x, y) divides the line segment joining A(x₁, y₁) and B(x₂, y₂) in ratio m:n (internally):

x = (mx₂ + nx₁) / (m + n) y = (my₂ + ny₁) / (m + n)

Midpoint (Special Case)

When P is the midpoint, m = n = 1: Midpoint = ((x₁+x₂)/2, (y₁+y₂)/2)

Examples

Example 1: Find point dividing line joining A(2, 3) and B(8, 9) in ratio 2:1.

  • x = (2·8 + 1·2)/3 = 18/3 = 6
  • y = (2·9 + 1·3)/3 = 21/3 = 7
  • Point: (6, 7)

Example 2: Find midpoint of A(4, −3) and B(8, 5).

  • ((4+8)/2, (−3+5)/2) = (6, 1)

5. Area of Triangle

Formula

For triangle with vertices A(x₁, y₁), B(x₂, y₂), C(x₃, y₃):

Area = (1/2) |x₁(y₂−y₃) + x₂(y₃−y₁) + x₃(y₁−y₂)|

Note

  • Use absolute value (area is always positive)
  • If area = 0, points are COLLINEAR (on same line)

Examples

Example 1: Find area of triangle with vertices A(1, 2), B(4, 5), C(6, 1).

  • Area = (1/2) |1(5−1) + 4(1−2) + 6(2−5)|
  • = (1/2) |4 − 4 − 18|
  • = (1/2)(18) = 9 sq units

Example 2: Check if A(2, 4), B(4, 6), C(6, 8) are collinear.

  • Area = (1/2) |2(6−8) + 4(8−4) + 6(4−6)|
  • = (1/2) |−4 + 16 − 12|
  • = (1/2)(0) = 0
  • YES, collinear.

6. Applications

Coordinate Geometry in Various Problems

Type 1: Equation of straight line

  • Slope = (y₂ − y₁) / (x₂ − x₁)
  • Line: y − y₁ = m(x − x₁)

Type 2: Parallel lines have EQUAL slopes Type 3: Perpendicular lines have slopes m₁ × m₂ = −1 Type 4: Find center of circle Type 5: Find centroid of triangle = ((x₁+x₂+x₃)/3, (y₁+y₂+y₃)/3)


7. Worked Examples

Example 1: Equilateral Triangle

Show that points (0, 0), (3, √3), and (3, −√3) form an equilateral triangle.

  • Side 1: from (0,0) to (3, √3) = √(9 + 3) = √12 = 2√3
  • Side 2: from (0,0) to (3, −√3) = √(9 + 3) = 2√3
  • Side 3: from (3, √3) to (3, −√3) = √(0 + 12) = 2√3
  • All sides equal = 2√3 → EQUILATERAL ✓

Example 2: Find Coordinates

Find coordinates of points trisecting line joining A(2, 1) and B(5, 8).

  • One trisecting point divides 1:2:
    • x = (1·5 + 2·2)/3 = 9/3 = 3
    • y = (1·8 + 2·1)/3 = 10/3
    • Point: (3, 10/3)
  • Other trisecting point divides 2:1:
    • x = (2·5 + 1·2)/3 = 12/3 = 4
    • y = (2·8 + 1·1)/3 = 17/3
    • Point: (4, 17/3)

Example 3: Centroid

Find centroid of triangle with vertices (0, 0), (6, 0), (3, 9).

  • Centroid = ((0+6+3)/3, (0+0+9)/3) = (3, 3)

Example 4: Circle Centre

A circle passes through (0, 0), (2, 0), and (0, 2). Find centre.

  • Centre is equidistant from all three points.
  • Let centre = (h, k). Distance to each = r.
  • h² + k² = (h−2)² + k² → 4h = 4 → h = 1
  • h² + k² = h² + (k−2)² → 4k = 4 → k = 1
  • Centre: (1, 1)

8. Common Mistakes

  1. Wrong coordinate order

    • Always (x, y) — first horizontal, then vertical.
  2. Forgetting absolute value in area

    • Use |...| to ensure positive area.
  3. Section formula in external division

    • For internal: (mx₂ + nx₁)/(m+n). For external (not in Class 10 syllabus normally): (mx₂ − nx₁)/(m−n).
  4. Mixing up internal and external ratios

    • 'Internal' means point INSIDE the segment.
  5. Wrong midpoint formula

    • Midpoint = ((x₁+x₂)/2, (y₁+y₂)/2). Sum first, then divide.

9. Indian Heritage

While Cartesian coordinates were formalised by Descartes (1637), Indian mathematicians used:

  • Bhaskara II (12th century): worked with curves and surfaces
  • Vedic geometry: precise altar geometry
  • Aryabhata (5th century): used coordinate-like thinking

Modern coordinate geometry combines Indian algebraic ideas with European geometric notation.


10. Conclusion

Coordinate Geometry bridges algebra and geometry:

  • Points become numerical pairs
  • Lines become equations
  • Geometric problems solved algebraically

Master:

  • Distance Formula
  • Section Formula
  • Area Formula

Foundation for:

  • Class 11 Conic Sections
  • Class 12 3D Geometry, Calculus
  • Engineering, Physics
  • Computer Graphics

The Cartesian Plane is the canvas on which all of mathematics is drawn.

Key formulas & results

Everything you need to memorise, in one card. Screenshot this for revision.

Distance
PQ = √((x₂−x₁)² + (y₂−y₁)²)
Section formula
x = (mx₂+nx₁)/(m+n); y = (my₂+ny₁)/(m+n)
Internal division
Midpoint
((x₁+x₂)/2, (y₁+y₂)/2)
Area of triangle
(1/2)|x₁(y₂−y₃) + x₂(y₃−y₁) + x₃(y₁−y₂)|
Centroid
((x₁+x₂+x₃)/3, (y₁+y₂+y₃)/3)
⚠️

Common mistakes & fixes

These are the exact errors that cost students marks in board exams. Read them once, save yourself the trouble.

WATCH OUT
Wrong coordinate order
Always (x, y) — horizontal first, then vertical.
WATCH OUT
Forgot absolute value for area
Area is always positive. Use |...|.

NCERT exercises (with solutions)

Every NCERT exercise from this chapter — what it covers and how many questions to expect.

Practice problems

Try each one yourself before tapping "Show solution". Active recall > rereading.

Q1EASY· Distance
Find the distance between (3, 4) and (6, 8).
Show solution
✦ Answer: d = √((6−3)² + (8−4)²) = √(9 + 16) = √25 = 5 units.
Q2MEDIUM· Section
Find the coordinates of the point dividing (2, 4) and (8, 10) in the ratio 2:1.
Show solution
Step 1 — Apply section formula. x = (mx₂ + nx₁)/(m+n) = (2(8) + 1(2))/(2+1) = 18/3 = 6 y = (my₂ + ny₁)/(m+n) = (2(10) + 1(4))/3 = 24/3 = 8 Step 2 — Verify. Distance from (2,4) to (6,8) = √(16+16) = √32 Distance from (6,8) to (8,10) = √(4+4) = √8 Ratio: √32 : √8 = 4√2 : 2√2 = 2:1 ✓ ✦ Answer: (6, 8).
Q3HARD· Application
Find the area of the triangle whose vertices are A(2,3), B(5,7), and C(8,3). Also check if it's right-angled.
Show solution
Step 1 — Area formula. Area = (1/2)|x₁(y₂−y₃) + x₂(y₃−y₁) + x₃(y₁−y₂)| = (1/2)|2(7−3) + 5(3−3) + 8(3−7)| = (1/2)|8 + 0 − 32| = (1/2)(24) = 12 sq units Step 2 — Check sides. AB = √((5−2)² + (7−3)²) = √(9+16) = 5 BC = √((8−5)² + (3−7)²) = √(9+16) = 5 AC = √((8−2)² + (3−3)²) = √36 = 6 Step 3 — Check right angle. Check Pythagoras: 5² + 5² = 50; 6² = 36. Not equal, so NOT right-angled by AB-BC. Try AB-AC: 5² + 6² = 61. BC² = 25. Not equal. Try BC-AC: 5² + 6² = 61. AB² = 25. Not equal. NO right angle. Step 4 — But isosceles! AB = BC = 5; AC = 6. Isosceles triangle. ✦ Answer: Area = 12 sq units. Triangle is ISOSCELES (AB = BC = 5) but NOT right-angled.

5-minute revision

The whole chapter, distilled. Read this the night before the exam.

  • Distance: √((x₂−x₁)² + (y₂−y₁)²)
  • Section formula: weighted average
  • Midpoint: simple average
  • Area: (1/2)|determinant-like sum|
  • Centroid: average of vertices
  • Collinear: area = 0
  • Descartes (1637) formalised coordinate geometry

CBSE marks blueprint

Where the marks come from in this chapter — so you can plan your prep.

Typical chapter weightage: 6-8 marks

Question typeMarks eachTypical countWhat it tests
MCQ12Formulas
Short Answer2-31-2Distance, section, midpoint
Long Answer50-1Area, collinearity, geometric problems
Prep strategy
  • Memorise all 4 formulas
  • Practice 15+ problems
  • Verify answers geometrically

Where this shows up in the real world

This chapter isn't just an exam topic — it lives in the world around you.

GPS and navigation

GPS uses coordinates to calculate distances.

Computer graphics

All graphics, video games use coordinate geometry.

Engineering CAD

Design software uses coordinates to model objects.

Exam strategy

Battle-tested tips from teachers and toppers for this chapter.

  1. Sketch problems first
  2. Use formulas precisely
  3. Verify with alternative method

Going beyond the textbook

For olympiad aspirants and curious learners — topics that build on this chapter.

  • Conic sections (Class 11)
  • 3D geometry
  • Vector geometry

Where else this chapter is tested

CBSE board isn't the only one — other exams test this chapter too.

CBSE Class 10 BoardVery High
Maths OlympiadHigh
JEEVery High

Questions students ask

The real ones — pulled from the Q&A community and tutor sessions.

Calculate area of triangle formed by them. If area = 0, points are collinear (on same straight line).
Verified by the tuition.in editorial team
Last reviewed on 20 May 2026. Written and reviewed by subject-matter experts — read about our process.
Editorial process →
Header Logo