Coordinate Geometry Chapter 7 | Mathematics

Complete NCERT notes, distance formula, section formula (internal and external), mid-point formula, area of triangle, collinearity, centroid, and key points for CBSE Class 10 Board Exam 2025-26.

📖 Introduction 📏 Distance Formula ✂️ Section Formula 🎯 Mid-Point Formula 📐 Area of Triangle 🔗 Collinearity ⚖️ Centroid 📐 All Formulas 📝 Examples ⭐ Key Points

📖 Introduction to Coordinate Geometry

Coordinate geometry (also known as analytic geometry) uses the Cartesian coordinate system to study geometric figures using algebraic equations. Points are represented as (x, y) on the plane.

📌 Cartesian Plane

X-axis (horizontal), Y-axis (vertical). Origin O = (0,0). Distance from origin: √(x²+y²)

📏 Distance Formula

📌 Formula

Distance between A(x₁, y₁) and B(x₂, y₂) = √[(x₂-x₁)² + (y₂-y₁)²]

📝 Example

Find distance between (2,3) and (5,7).
d = √[(5-2)² + (7-3)²] = √[3²+4²] = √25 = 5 units

💡 Distance from origin: √(x² + y²)

✂️ Section Formula (Internal Division)

📌 Formula

If point P divides AB in ratio m₁ : m₂, then coordinates of P are:

P = ((m₁x₂ + m₂x₁)/(m₁+m₂), (m₁y₂ + m₂y₁)/(m₁+m₂))

📝 Example

Find coordinates of point dividing (2,3) and (4,5) in ratio 1:2 internally.
x = (1×4 + 2×2)/(1+2) = (4+4)/3 = 8/3
y = (1×5 + 2×3)/(3) = (5+6)/3 = 11/3
Point = (8/3, 11/3)

💡 External Division: P = ((m₁x₂ - m₂x₁)/(m₁-m₂), (m₁y₂ - m₂y₁)/(m₁-m₂))

🎯 Mid-Point Formula

📌 Formula (Special case of section formula when m₁ = m₂)

Mid-point of A(x₁, y₁) and B(x₂, y₂) = ((x₁+x₂)/2, (y₁+y₂)/2)

📐 Area of Triangle

📌 Formula

Area = ½ |x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂)|

The absolute value ensures positive area.

📝 Example

Find area of triangle with vertices (0,0), (4,0), (0,3).
Area = ½ |0(0-3) + 4(3-0) + 0(0-0)| = ½ |12| = 6 sq units

🔗 Condition for Collinearity

Three points A, B, C are collinear if area of triangle ABC = 0.

x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂) = 0

💡 Alternatively, slope AB = slope BC.

⚖️ Centroid of a Triangle

📌 Formula

Centroid G = ((x₁+x₂+x₃)/3, (y₁+y₂+y₃)/3)

The centroid divides the medians in the ratio 2:1.

📐 All Important Formulas

1️⃣ Distance Formula

d = √[(x₂-x₁)² + (y₂-y₁)²]

2️⃣ Section Formula (Internal)

((m₁x₂+m₂x₁)/(m₁+m₂), (m₁y₂+m₂y₁)/(m₁+m₂))

3️⃣ Mid-Point Formula

((x₁+x₂)/2, (y₁+y₂)/2)

4️⃣ Area of Triangle

½|x₁(y₂-y₃)+x₂(y₃-y₁)+x₃(y₁-y₂)|

5️⃣ Collinearity Condition

x₁(y₂-y₃)+x₂(y₃-y₁)+x₃(y₁-y₂)=0

6️⃣ Centroid

((x₁+x₂+x₃)/3, (y₁+y₂+y₃)/3)

📝 NCERT Solved Examples

Example 1: Distance Formula

Show that points (1,2), (3,4), (5,6) are collinear.
AB = √[(3-1)²+(4-2)²] = √(4+4)=√8
BC = √[(5-3)²+(6-4)²] = √(4+4)=√8
AC = √[(5-1)²+(6-2)²] = √(16+16)=√32=2√8
Since AB+BC=AC, points are collinear.

Example 2: Section Formula

Find coordinates of point dividing (4,5) and (7,8) in ratio 2:3 internally.
x = (2×7 + 3×4)/5 = (14+12)/5 = 26/5
y = (2×8 + 3×5)/5 = (16+15)/5 = 31/5
Point = (26/5, 31/5)

Example 3: Area of Triangle

Find area of triangle with vertices (2,3), (4,5), (6,1).
Area = ½ |2(5-1) + 4(1-3) + 6(3-5)|
= ½ |2×4 + 4×(-2) + 6×(-2)| = ½ |8 -8 -12| = ½ × 12 = 6 sq units

Example 4: Centroid

Find centroid of triangle with vertices (2,3), (4,5), (6,7).
G = ((2+4+6)/3, (3+5+7)/3) = (12/3, 15/3) = (4,5)

⭐ Key Points for Board Exam

Distance formula is derived from Pythagoras theorem.
Section formula can be remembered as weighted average.
Mid-point is average of coordinates.
Area of triangle formula works for any three non-collinear points.
If area = 0 → points are collinear.
Centroid divides median in ratio 2:1 from vertex to opposite side.
Coordinates of any point on x-axis: (x,0); on y-axis: (0,y).
Distance from origin: √(x²+y²).

📌 Quick Reference Table

Concept	Formula
Distance欧	√[(x₂-x₁)²+(y₂-y₁)²]
Section (internal)	((m₁x₂+m₂x₁)/(m₁+m₂), (m₁y₂+m₂y₁)/(m₁+m₂))
Mid-point	((x₁+x₂)/2, (y₁+y₂)/2)
Area of triangle	½\|x₁(y₂-y₃)+x₂(y₃-y₁)+x₃(y₁-y₂)\|
Centroid	((x₁+x₂+x₃)/3, (y₁+y₂+y₃)/3)

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