x y A(x₁,y₁) B(x₂,y₂) C(x₃,y₃) mid m:n
(x,y)
Chapter 7  ·  Class X Mathematics

Algebra Meets Geometry on the Cartesian Plane

Coordinate Geometry

Distance, Midpoint, Area — Three Formulae, Infinite Problems

📖 Read Full Notes ⚡ Formula Capsules

Chapter Snapshot

7Concepts
6Formulae
8–10%Exam Weight
4–5Avg Q's
ModerateDifficulty

Why This Chapter Matters for Exams

CBSE BoardNTSEState Boards

Coordinate Geometry consistently delivers 8–10 marks in CBSE Boards. The Section Formula and Area of Triangle Formula are high-frequency question types. Finding centroids, determining collinearity, and proving quadrilateral types using distance formula are standard 3–4 mark problems.

Key Concept Highlights

Distance Formula
Section Formula (Internal)
Section Formula (External)
Midpoint Formula
Area of a Triangle
Centroid of a Triangle
Collinearity of Three Points
Nature of Quadrilateral via Distance

Important Formula Capsules

$PQ = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$
$\text{Section (m:n internal): }\left(\tfrac{mx_2+nx_1}{m+n},\ \tfrac{my_2+ny_1}{m+n}\right)$
$\text{Midpoint: }\left(\tfrac{x_1+x_2}{2},\ \tfrac{y_1+y_2}{2}\right)$
$\text{Area of }\triangle = \tfrac{1}{2}|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|$
$\text{Centroid: }\left(\tfrac{x_1+x_2+x_3}{3},\ \tfrac{y_1+y_2+y_3}{3}\right)$
$\text{Collinear if Area of }\triangle = 0$

What You Will Learn

Navigate to Chapter Resources

📖 Full Notes Complete NCERT Coverage 🧠 MCQ Practice Topic-wise Questions 📘✔️ Exercises Step-by-step NCERT Solutions ✔️❌ True / False Concept-based True & False

🏆 Exam Strategy & Preparation Tips

These three formulae (Distance, Section, Area) are the entire chapter — master them completely. Practice "show that ABCD is a rhombus/parallelogram/square" problems systematically. Collinearity via area = 0 is a quick 2-mark answer. Time investment: 2 days.

Chapter 7 · CBSE Class X
📍
Coordinate Geometry
Distance Formula & Section Formula
Coordinate Geometry CBSE Class X
📘 Definition

Definition

  • Horizontal axis → x-axis
  • Vertical axis → y-axis
  • Vertical axis → y-axis
🎨 SVG Diagram

Graphical Representation

(0,0) (3,4) X Y
💡 Concept

Key Concepts

  • Ordered Pair (x, y): Represents position of a point
  • Abscissa (x-coordinate): Distance from y-axis
  • Ordinate (y-coordinate): Distance from x-axis
  • Quadrants: Plane divided into 4 regions based on signs of coordinates
💡 Concept

Quadrants and Sign Convention

🔢 Formula

Important Formulae

Distacne Formula
\[ \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
Mid-ponit Formula
\[ \left(\dfrac{x_1 + x_2}{2}\right) , \left(\dfrac{y_1 + y_2}{2}\right) \]
📄 Insight
The distance formula is derived using the Pythagoras theorem by forming a right triangle between two points. Horizontal and vertical differences act as base and height respectively.
✏️ Example

Solved Examples

Example 1: Identify quadrant of point (5, −3)

Since x is positive and y is negative → Quadrant IV

Example 2: Identify quadrant of point (−7, 2)

Since x is negative and y is positive → Quadrant II

Example 3: Where does point (0, 4) lie?

On y-axis (not in any quadrant)

📐 Derivation

Logical Derivation of Quadrants

The axes split the plane into four parts based on sign changes: Moving right increases x positively, moving left decreases x negatively. Similarly, moving up increases y positively and moving down decreases y negatively. Combining these gives the four sign regions (quadrants).
⚡ Exam Tip

Exam Tips for Board Aspirants

  • Exam Tips for Board Aspirants
  • Always check sign before answering quadrant-based MCQs
  • Coordinate geometry questions frequently involve quadrant identification
  • Graph-based case study questions are common in CBSE exams
⚠️ Warning

Common Mistakes

  • Confusing Quadrant II and III
  • Ignoring negative signs while plotting
  • Considering points on axes as part of quadrants
📋 Case Study
A triangle has vertices A(2,3), B(-4,5), and C(-3,-6).
Identify the quadrants of each vertex and determine if the triangle spans all four quadrants.
Solution Insight:
A → Q1, B → Q2, C → Q3 → Triangle spans three quadrants (not all four)
🌟 Importance

Why This Topic is Important

Maths Dept · Updated 2026
📍
Distance Formula
📘 Definition

Distance Formula in Coordinate Geometry – Complete Board Guide

🔢 Formula

Distance between two points (x₁, y₁) and (x₂, y₂):

\[ \sqrt{(x_2 − x_1)^2 + (y_2 − y_1)^2} \]
🎨 SVG Diagram

Geometric Visualization

A(x₁,y₁) B(x₂,y₂) Base (Δx) Height (Δy) Distance
🔬 Proof

Derivation Using Pythagoras Theorem

Consider two points A(x₁, y₁) and B(x₂, y₂). Draw a right triangle by projecting horizontal and vertical lines.
  • Horizontal side = |x₂ − x₁|
  • Vertical side = |y₂ − y₁|
  • Applying Pythagoras theorem:
\[ \begin{aligned}\text{Distance}^2 = (x_2 − x_1)^2 + (y_2 − y_1)^2 \\\\ \text{Distance} = \sqrt{(x_2 − x_1)^2 + (y_2 − y_1)^2}\end{aligned} \]
📌 Note

Consider two points A(x₁, y₁) and B(x₂, y₂). Draw a right triangle by projecting horizontal and vertical lines.

  • Proving triangles are isosceles or equilateral
  • Verifying right-angled triangle using Pythagoras
  • Checking collinearity of three points
  • Finding perimeter of triangle or polygon
✏️ Example

Solved Examples

Example-1

Find distance between A(2,3) and B(6,7)

\[ \begin{aligned} \text{Distance} &= \sqrt{(6-2)^2 + (7-3)^2} \\ &= \sqrt{16 + 16} \\ &= \sqrt{32} \\ &= 4\sqrt{2} \end{aligned} \]

Example-2

Determine if triangle with points A(0,0), B(3,4), C(6,8) is a straight line

\[ \begin{aligned} AB &= 5,\ BC = 5,\ AC = 10 \\ \Rightarrow\ &AB + BC = AC \\ \Rightarrow\ &\text{Points are collinear} \end{aligned} \]
⚡ Exam Tip

Exam Tips for Board Aspirants

  • Always simplify square roots properly (e.g., \(\sqrt{32} = 4\sqrt{2}\))
  • Use distance formula before applying triangle properties
  • For right triangle: verify \(a^2 + b^2 = c^2\)
  • Keep values in squared form when comparing distances to save time
⚠️ Warning

Common Mistakes

  • Forgetting to square differences
  • Mixing up x and y coordinates
  • Incorrect simplification of square roots
📋 Case Study

CBSE Case Study (HOTS)

Points A(1,2), B(4,6), and C(6,2) form a triangle. Verify whether it is right-angled.

Strategy: Find all three side lengths using distance formula and check Pythagoras relation.

🌟 Importance

Why This Topic is Important

📍
Distance of a Point from the Origin
📘 Definition

Distance of a Point from the Origin

🔢 Formula

Formula

\[ \text{Distance} = \sqrt{(x^2 + y^2)} \]
This is a direct application of the distance formula by taking one point as the origin.
🎨 SVG Diagram

Geometric Visualization

(0,0) (x,y) x y Distance
📐 Derivation

Derivation

Consider a point P(x, y) and the origin O(0, 0). A right triangle is formed with:

  • Base = x units
  • Height = y units

Using Pythagoras theorem:

  • \[OP^2 = x^2 + y^2\]
    • Therefore,
  • \[OP = \sqrt{x^2 + y^2}\]

✏️ Example

Solved Example

Example-1

Find distance of point (3,4) from origin
\[ \begin{aligned}\text{Distance} &= \sqrt{(3^2 + 4^2)}\\ & = \sqrt{(9 + 16)}\\ &= \sqrt{25} \\ &= 5\end{aligned} \]
✏️ Example

Example-2

Find distance of point (-5,12) from origin
\[ \begin{aligned}\text{Distance} &= \sqrt{((-5)^2 + 12^2)}\\ & = \sqrt{(25 + 144)}\\ &= \sqrt{169} \\ &= 13\end{aligned} \]
📌 Note

Applications in Triangles Chapter

  • Finding radius when origin is center of circle
  • Verifying symmetry of points about origin
  • Simplifying triangle side calculations when one vertex is origin
⚡ Exam Tip

Exam Tips

  • Always square both x and y (negative signs disappear)
  • Recognize common Pythagorean triples (3-4-5, 5-12-13)
  • Use this shortcut when one point is origin to save time
⚠️ Warning

Common Mistakes

  • Forgetting to square negative values
  • Writing \(\sqrt{(x + y²)}\) instead of \(\sqrt{(x² + y²)}\)
  • Not simplifying square roots
📋 Case Study

CBSE Case Study (HOTS)

A point P lies at a distance of 10 units from the origin and has x-coordinate 6. Find its y-coordinate.

Strategy: Use

\[ \begin{aligned}\sqrt{(x^2 + y^2)} &= 10 \\\Rightarrow 36 + y^2 &= 100 \Rightarrow y^2\\ &= 64 \Rightarrow y \\&= ±8\end{aligned} \]
🌟 Importance

Why This Topic is Important

📍
Examples
❓ Question

Verifying Triangle and Identifying Its Type

Example-1

Question

Do the points (3, 2), (−2, −3) and (2, 3) form a triangle? If yes, identify the type of triangle.

🎨 SVG Diagram

Geometric Visualization

A(3,2) B(-2,-3) C(2,3)
📄 Solution
  1. Let A(3,2), B(−2,−3), C(2,3)
  2. Find AB
  3. \begin{aligned} AB &= \sqrt{[(3 − (−2))2 + (2 − (−3))2]}\\ &= \sqrt{[(5)^2 + (5)^2]} \\&= \sqrt{(25 + 25)} \\&= \sqrt{50}\end{aligned}
  4. Find AB
  5. \begin{aligned}BC& = \sqrt{[(2 − (−2))² + (3 − (−3))²]} \\ &= \sqrt{[(4)² + (6)²]} \\&= \sqrt{(16 + 36)} \\&= \sqrt{52}\end{aligned}
  6. Find CA
  7. \begin{aligned}CA &= \sqrt{[(3 − 2)² + (2 − 3)²]} \\ &= \sqrt{[(1)² + (−1)²]} \\&= \sqrt{(1 + 1)} \\&= \sqrt{2}\end{aligned}

  8. Check if Points Form a Triangle

  9. Sum of any two sides must be greater than the third:
  10. \[\sqrt{50} + \sqrt{2} > \sqrt{52}\]
  11. \[\sqrt{52} + \sqrt{2} > \sqrt{50}\]
  12. \[\sqrt{50} + \sqrt{52} > \sqrt{2}\]

  13. All conditions are satisfied → Points form a triangle

📌 Note

Identify Type of Triangle

  1. Check Pythagoras theorem:
  2. \begin{aligned}AB^2 + CA^2 = \left(\sqrt{50}\right)^2 + \left(\sqrt{2}\right)^2 = 50 + 2 = 52 = \left(\sqrt{52}\right)^2 = BC^2\end{aligned}
  3. Since Pythagoras theorem is satisfied, triangle ABC is a Right-Angled Triangle.
⚡ Exam Tip

Exam Tips

  • Always compare squares (avoid unnecessary square roots)
  • Use Pythagoras for type identification
  • Verify triangle existence before classifying
⚠️ Warning

Common Mistakes

  • Incorrect expansion of (2 − (−3))²
  • Wrong substitution in BC calculation
  • Typo using 'z' instead of '2' in coordinates
  • Skipping triangle inequality verification
📋 Case Study

CBSE HOTS Extension

Find the coordinates of a point D such that ABCD forms a rectangle.

Hint: Use vector or midpoint method → D = A + C − B

🌟 Importance

Why This Example is Important

📍
Examples
❓ Question

Proving Points Form a Square

Example-2

Question

Show that the points (1, 7), (4, 2), (−1, −1) and (−4, 4) are the vertices of a square.

🎨 SVG Diagram

Geometric Visualization

A(1,7) B(4,2) C(-1,-1) D(-4,4)
🧩 Solution

Step-by-Step Solution

  1. Let A(1,7), B(4,2), C(−1,−1), D(−4,4)

  2. Find All Side Lengths

  3. \begin{aligned}AB^2 &= (4−1)^2 + (2−7)^2 \\&= 3^2 + (−5)^2 \\&= 9 + 25 \\&= 34\end{aligned}
  4. \begin{aligned}BC^2 &= (−1−4)^2 + (−1−2)^2 \\&= (−5)^2 + (−3)^2 \\&= 25 + 9 \\&= 34\end{aligned}
  5. \begin{aligned}CD^2 &= (−4+1)^2 + (4+1)^2 \\&= (−3)^2 + (5)^2 \\&= 9 + 25 \\&= 34\end{aligned}
  6. \begin{aligned}DA^2 &= (1+4)^2 + (7−4)^2 \\&= (5)^2 + (3)^2 \\&= 25 + 9 \\&= 34\end{aligned}
  7. \[\Rightarrow AB = BC = CD = DA \Rightarrow\text{ All sides are equal}\]

  8. Check Diagonals

  9. \begin{aligned}AC^2 &= (−1−1)^2 + (−1−7)^2 \\&= (−2)^2 + (−8)^2 \\&= 4 + 64 \\&= 68\end{aligned}
  10. \begin{aligned}BD^2 &= (−4−4)^2 + (4−2)^2 \\&= (−8)^2 + (2)^2 \\&= 64 + 4 \\&= 68\end{aligned}
  11. \[\Rightarrow AC = BD \Rightarrow \text{ Diagonals are equal}\]

  12. Verify Right Angle (Key Condition)

  13. \begin{aligned}AB² + BC² &= 34 + 34 \\&= 68 \\&= AC²\end{aligned}
  14. \[\Rightarrow \angle ABC = 90^\circ\]

  15. Final Conclusion

  16. All sides are equal
  17. Diagonals are equal
  18. One angle is 90°
  19. Therefore, ABCD is a Square.
⚡ Exam Tip

Exam Tips

  • Always compare squares to save time (avoid roots)
  • For square: check equal sides + right angle OR equal diagonals
  • Maintain consistent point order (A → B → C → D)
⚠️ Warning

Common Mistakes

  • Wrong coordinate of point D (should be −4,4 not −4,−4)
  • Incorrect formula expansion in multiple steps
  • Not verifying right angle condition
  • Random algebra instead of structured distance calculation
📋 Case Study

CBSE HOTS Extension

Find the area of the square using coordinate geometry.

Hint: Side = \(\sqrt{34}\) → Area = 34 square units

🌟 Importance

Why This Example is Important

📍
Examples
❓ Question

Locus of Points Equidistant from Two Points

Example-3

Question

Find a relation between x and y such that the point (x, y) is equidistant from the points (7, 1) and (3, 5).

🎨 SVG Diagram

Geometric Visualization

A(7,1) B(3,5) M
💡 Concept

Concept Insight

🧩 Solution

Step-by-Step Solution

  1. Let point P(x, y) be equidistant from A(7,1) and B(3,5)

  2. Apply Distance Formula

  3. \[\sqrt{(x−7)^2 + (y−1)^2} = \sqrt{(x−3)^2 + (y−5)^2}\]
  4. \[(x−7)^2 + (y−1)^2 = (x−3)^2 + (y−5)^2\]
  5. \[x^2 −14x +49 + y^2 −2y +1 = x^2 −6x +9 + y^2 −10y +25\]

  6. Simplify

  7. \[−14x −2y +50 = −6x −10y +34\]
  8. \[−14x +6x −2y +10y = 34 −50\]
  9. \[−14x +6x −2y +10y = 34 −50\]
  10. Divide by −8:
  11. \[x − y = 2\]

  12. Final Answer

  13. Required relation: x − y = 2
📌 Note

Geometric Interpretation

⚡ Exam Tip

Exam Tips

  • Always square both sides to remove root safely
  • Cancel x² and y² early to simplify quickly
  • Final answer should be linear (straight line equation)
⚠️ Warning

Common Mistakes

📋 Case Study

CBSE HOTS Extension

Find coordinates of the midpoint of (7,1) and (3,5), and verify that it satisfies the equation x − y = 2.
🌟 Importance

Why This Example is Important

📍
Examples
❓ Question

Locus of Points Equidistant from Two Points

Example-4

Question

Find a relation between x and y such that the point (x, y) is equidistant from the points (7, 1) and (3, 5).

💡 Concept

Concept Insight

🎨 SVG Diagram

Geometric Visualization

A(7,1) B(3,5) M
🧩 Solution

Step by step Solution

  1. Apply Distance Formula

  2. \[\sqrt{[(x−7)^2 + (y−1)^2]} = \sqrt{[(x−3)^2 + (y−5)^2]}\]

  3. Squaring both side to Remove Square Root

  4. \[(x−7)^2 + (y−1)^2 = (x−3)^2 + (y−5)^2\]
  5. \[x^2 −14x +49 + y^2 −2y +1 = x^2 −6x +9 + y^2 −10y +25\]

  6. Step 4: Simplify

  7. \[−14x −2y +50 = −6x −10y +34\]
  8. \[−14x +6x −2y +10y = 34 −50\]
  9. \[−8x +8y = −16\]
  10. Divide by −8:\]
  11. \[x − y = 2\]

  12. Final Answer

  13. Required relation: \[x − y = 2\]
📌 Note

Geometric Interpretation

⚡ Exam Tip

Exam Tips

  • Always square both sides to remove root safely
  • Cancel x² and y² early to simplify quickly
  • Final answer should be linear (straight line equation)
⚠️ Warning

Common Mistakes

  • Incorrect expansion of (y−1)² and (y−5)²
  • Sign errors while simplifying
  • Not reducing to linear equation
📋 Case Study

CBSE HOTS Extension

Find coordinates of the midpoint of (7,1) and (3,5), and verify that it satisfies the equation x − y = 2.

🌟 Importance

Why This Example is Important

📍
Section Formula (Internal Division)
📄 Defition

Definition

The Section Formula is used to find the coordinates of a point that divides a line segment joining two given points in a specific ratio. When the point lies between the two points, it is called internal division.
🔢 Formula

Formula

\[ \begin{aligned}x = (m x₂ + n x₁) / (m + n)\\ y = (m y₂ + n y₁) / (m + n)\end{aligned} \]
🎨 SVG Diagram

Geometric Visualization

A(x₁,y₁) B(x₂,y₂) P(x,y) n m
📐 Derivation

Derivation

Assume point P divides AB internally in the ratio m:n. Using weighted averages of coordinates:

  • x-coordinate is weighted average of x₁ and x₂
  • y-coordinate is weighted average of y₁ and y₂

Hence, coordinates shift towards the point with larger weight.

✏️ Example

Solved Example

Find the point dividing A(2,4) and B(8,10) in the ratio 1:2 internally

\[\begin{aligned}x &= (1×8 + 2×2)/(1+2) \\&= (8 + 4)/3 \\&= 4 \\\\ y &= (1×10 + 2×4)/(1+2) \\&= (10 + 8)/3 \\&= 6\end{aligned}\]

Required point = (4, 6)

📌 Note

Applications

  • Finding coordinates of centroid of triangle
  • Dividing a line segment in a given ratio
  • Coordinate proofs in triangles and polygons
  • Solving locus and midpoint related problems
⚡ Exam Tip

Exam Tips

  • Remember: m is multiplied with second point (x₂, y₂)
  • Always write ratio in correct order (AP:PB = m:n)
  • Use shortcut: bigger weight pulls point closer
⚠️ Warning

Common Mistakes

  • Interchanging m and n incorrectly
  • Interchanging m and n incorrectly
  • Forgetting denominator (m+n)
📋 Case Study

CBSE Case Study (HOTS)

Find the coordinates of the centroid of triangle with vertices A(1,2), B(4,6), C(7,8).

Hint: Use section formula repeatedly or apply centroid formula.

🌟 Importance

Why This Topic is Important

📍
Midpoint Formula
📄 Defition

Definition

The midpoint formula is used to find the exact middle point of a line segment joining two points in the Cartesian plane. It divides the line segment into two equal parts.
🔢 Formula

Formula

\[ \begin{aligned}M = ((x₁ + x₂)/2 , (y₁ + y₂)/2)\end{aligned} \]
🎨 SVG Diagram

Geometric Visualization

A(x₁,y₁) B(x₂,y₂) M Equal Distance Equal Distance
📐 Derivation

Derivation

The midpoint divides the line segment in the ratio 1:1 internally.

Using section formula:

\[ \begin{aligned}x &= (1·x₂ + 1·x₁) / (1+1) \\&= (x₁ + x₂)/2 \\\\ y &= (1·y₂ + 1·y₁) / (1+1) \\&= (y₁ + y₂)/2\end{aligned} \]
✏️ Example

Solved Example

Find midpoint of A(2,4) and B(6,8)

\[ M = ((2+6)/2 , (4+8)/2) = (8/2 , 12/2) = (4 , 6) \]
📌 Note

Applications

  • Finding centroid and medians of triangle
  • Verifying symmetry about a point
  • Proving parallelogram properties (diagonals bisect each other)
  • Used in coordinate proofs of geometric figures
⚡ Exam Tip

Exam Tips

  • Add coordinates first, then divide (avoid calculation mistakes)
  • Useful shortcut when midpoint is origin → x₁ + x₂ = 0, y₁ + y₂ = 0
  • Frequently used in case-study and MCQs
⚠️ Warning

Common Mistakes

  • Forgetting to divide by 2
  • Mixing x and y coordinates
  • Arithmetic errors in addition
📋 Case Study

CBSE Case Study (HOTS)

The midpoint of line joining A(2,3) and B(x,7) is (4,5). Find x.

Strategy: (2 + x)/2 = 4 → x = 6

🌟 Importance

Why This Topic is Important

📍
Examples
❓ Question

Using Section Formula (Internal Division)

Example-5

Question

Find the coordinates of the point which divides the line segment joining the points (4, −3) and (8, 5) in the ratio 3 : 1 internally.

🎨 SVG Diagram

Geometric Visualization

A(4,-3) B(8,5) P(x,y) 1 3
🧩 Solution

Step by step Solution

  1. Let A(4, −3) and B(8, 5)

  2. \[\sqrt{[(x−7)^2 + (y−1)^2]} = \sqrt{[(x−3)^2 + (y−5)^2]}\]

  3. Given ratio AP : PB = 3 : 1 → m = 3, n = 1

  4. Apply Section Formula

  5. \[\begin{aligned}x &= (m·x₂ + n·x₁) / (m + n)\\ &= (3×8 + 1×4) / (3+1) \\&= (24 + 4)/4 \\&= 28/4 \\&= 7\end{aligned}\]
  6. \[\begin{aligned}y &= (m·y₂ + n·y₁) / (m + n)\\ &= (3×5 + 1×(−3)) / 4 \\&= (15 − 3)/4 \\&= 12/4 \\&= 3\end{aligned}\]

  7. Final Answer

  8. Required point \[P = (7, 3)\]
💡 Concept

Concept Insight

⚡ Exam Tip

Exam Tips

  • Always multiply m with second point (x₂, y₂)
  • Keep track of negative signs carefully
  • Larger ratio value indicates closer point
⚠️ Warning

Common Mistakes

  • Writing denominator incorrectly (m₁+m₂ instead of m+n)
  • Swapping m and n during substitution
  • Ignoring negative sign in y-coordinate
📋 Case Study

CBSE HOTS Extension

Find the ratio in which point (7,3) divides the line joining (4,−3) and (8,5).
🌟 Importance

Why This Example is Important

📍
Practical Interpretation of Coordinate Geometry
📄 Body

Real World Applications

Concept Overview

Coordinate Geometry transforms geometric shapes into algebraic form using coordinates. This allows precise calculation, modeling, and visualization of real-world systems.

Real-Life Visualization

Location A Location B Distance / Route

Major Real-World Applications

  • GPS & Navigation Systems: Used to locate exact positions on Earth using latitude and longitude coordinates.
  • Computer Graphics & Animation: Every object in games, movies, and UI design is positioned using coordinate systems.
  • Engineering & Architecture: Blueprints and structural designs rely on coordinate-based precision.
  • Robotics & AI Navigation: Robots use coordinate grids to move, detect obstacles, and plan paths.
  • Astronomy: Positions of stars, planets, and satellites are mapped using coordinate systems.

Deep Insight (Why It Matters)

Coordinate Geometry acts as a bridge between algebra and geometry. It allows complex shapes and movements to be represented numerically, enabling precise calculations in science, technology, and real-life problem solving.

Real-Life Example

Suppose a delivery app needs to find the shortest route between two locations. It uses the distance formula to compute the minimum distance between two coordinate points, ensuring efficient navigation.

Common Mistakes

  • Treating coordinate geometry as purely theoretical (ignoring applications)
  • Not linking formulas to real-world use cases
  • Memorizing without understanding practical meaning

CBSE Case Study (HOTS)

A drone travels from point A(2,3) to B(10,11). Find the shortest distance traveled and explain how coordinate geometry helps in navigation.

Why This Topic is Important

This topic highlights the practical power of coordinate geometry. It strengthens conceptual understanding and prepares students for real-world problem solving, making it highly valuable beyond board examinations.

📍
Common Mistakes in Coordinate Geometry & How to Avoid Them
⚠️ Warning

Common Mistakes in Coordinate Geometry & How to Avoid Them

📍
Collinearity of Points
💡 Concept

Collinearity of Points – Complete Concept

📍
Area of Triangle
💡 Concept

Area of Triangle Using Coordinates

📍
Types of Triangles
💡 Concept

Types of Triangles Using Coordinates

📍
Centroid of Triangle
💡 Concept

Centroid of Triangle

📍
Centroid of Triangle
🌟 Importance

Important Shortcuts & Tricks

NCERT · Class X · Chapter 7
Coordinate Geometry

Master Distance, Section, Midpoint, Collinearity, Triangle Classification & Centroid — with an AI-powered solver, concept quizzes, and interactive visualisers.

📐

Key Formulas at a Glance

Distance Formula
\[d =\sqrt{[(x₂−x₁)² + (y₂−y₁)²]}\]
Distance from origin: \(\sqrt{(x² + y²)}\)
Section Formula (Internal)
\[P = \left( \dfrac{mx₂+nx₁}{m+n} ,\; \dfrac{my₂+ny₁}{m+n} \right)\]
Point P divides AB in ratio m : n internally
Midpoint Formula
\[M = \left( \dfrac{x₁+x₂}{2} ,\; \dfrac{y₁+y₂}{2}\right)\]
Special case: m = n = 1 in section formula
Centroid of Triangle
\[G = \left(\dfrac{x₁+x₂+x₃}{3} ,\; \dfrac{y₁+y₂+y₃}{3}\right)\]
Intersection point of all three medians
Area of Triangle
\[A = \dfrac{1}{2} \Big| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \Big|\]
Use |absolute value| — area is always positive
Collinearity Condition
\[\text{Area of △ABC }= 0\]
Alternatively: AB + BC = AC (if B is between A and C)
Section Formula (External)
\[P = \left(\dfrac{mx₂−nx₁}{m−n} ,\; \dfrac{my₂−ny₁}{m−n} \right)\]
m ≠ n; external division beyond segment AB
Right Triangle Check
\[a² + b² = c²\]
c = longest side; compute all three distances first
🤖

AI Step-by-Step Solver

Enter coordinates and choose an operation — the solver shows every working step.

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Tricks & Tips

  • 🔑Zero-ratio trick: When a point lies on the x-axis, its y-coordinate = 0. Use this to find the ratio in which the x-axis divides a line segment — set y = 0 in the section formula.
  • 📌Midpoint shortcut: If you know midpoint M and one end A, find the other end B: B = (2×Mx − x₁, 2×My − y₁).
  • ⚡Equilateral triangle: All three sides are equal. After computing the three distances, check AB = BC = CA — no need for Pythagoras.
  • 🎯Collinearity fast check: Compute area using the shoelace formula. If area = 0, the points are collinear — no need to find individual distances.
  • 📐Diagonals bisect ↔ parallelogram: If midpoints of both diagonals are the same, the quadrilateral is a parallelogram.
  • 🔄Negative ratio = external division: If section formula gives a negative value for m or n, the division is external.
  • 🏁Square vs Rhombus: Square has equal sides AND equal diagonals; rhombus has equal sides but unequal diagonals. Always compute diagonals to distinguish.
  • 🧮Area with fractions: Don't cancel midway — compute the entire numerator first, then apply ½. Fewer arithmetic errors.
⚠️

Common Mistakes & How to Avoid Them

  • ❌Forgetting the absolute value in area: The shoelace formula can give a negative result. Always take |…| to ensure the area is positive.
  • ❌Swapping m and n in section formula: The formula is (mx₂ + nx₁)/(m+n). The numerator uses x₂ with m (closer end), not x₁.
  • ❌Missing √ in distance formula: AB² = … is a squared distance. If a question asks for the distance (not distance²), always take the square root.
  • ❌Using wrong vertices for Pythagoras: First identify the longest side (hypotenuse), then verify c² = a² + b². Applying to the wrong pair gives a wrong classification.
  • ❌Not simplifying the ratio: Always express the ratio m : n in its simplest form. If you find the ratio as 4 : 2, write 2 : 1.
  • ❌Centroid ≠ circumcentre: The centroid formula G = ((Σx)/3,(Σy)/3) is only for centroid. Circumcentre requires a different method (perpendicular bisectors).
  • ❌Sign errors with negative coordinates: (−3)² = 9, not −9. Square before subtracting, especially with negative coordinates.
📝

Concept-Building Questions & Solutions

Original questions organised by concept — click to reveal the full step-by-step solution.

📏 Concept 1 — Distance Formula
Q1. Prove that the points A(−2, 5), B(3, −1) and C(7, 3) form a right-angled isosceles triangle. Medium▼
1
Compute all three side lengths using the distance formula.
2
AB = √[(3−(−2))² + (−1−5)²] = √[25+36] = √61
3
BC = √[(7−3)² + (3−(−1))²] = √[16+16] = √32 = 4√2
4
CA = √[(−2−7)² + (5−3)²] = √[81+4] = √85
5
For a right angle at B, check: AB² + BC² = 61 + 32 = 93 ≠ 85. Try at C: BC² + CA² = 32 + 85 = 117 ≠ 61. Try at A: AB² + CA² = 61 + 85 = 146 ≠ 32. So not right-angled. Verify equal sides: AB ≠ BC ≠ CA, so not isosceles either. (The question as stated has no solution — try modified version below.)
6
Modified: A(0,0), B(2,2), C(2,0). AB=√8, BC=2, CA=2. BC=CA → isosceles. BC²+CA²=4+4=8=AB² → right angle at C. ✓
∴ A(0,0), B(2,2), C(2,0) form a right-angled isosceles triangle (right angle at C).
Q2. Find the value of y if the distance between P(3, y) and Q(−1, 4) is 5 units. Easy▼
1
Apply distance formula: PQ = √[(−1−3)² + (4−y)²] = 5
2
Square both sides: (−4)² + (4−y)² = 25 → 16 + (4−y)² = 25
3
(4−y)² = 9 → 4−y = ±3
4
Case 1: 4−y = 3 → y = 1. Case 2: 4−y = −3 → y = 7.
∴ y = 1 or y = 7 (two valid values).
Q3. Show that the points A(1, 7), B(4, 2), C(−1, −1), D(−4, 4) are the vertices of a square. Hard▼
1
Compute all four sides. AB=√[(4−1)²+(2−7)²]=√[9+25]=√34
2
BC=√[(−1−4)²+(−1−2)²]=√[25+9]=√34
3
CD=√[(−4+1)²+(4+1)²]=√[9+25]=√34
4
DA=√[(1+4)²+(7−4)²]=√[25+9]=√34. All sides equal ✓
5
Compute diagonals: AC=√[(−1−1)²+(−1−7)²]=√[4+64]=√68
6
BD=√[(−4−4)²+(4−2)²]=√[64+4]=√68. Equal diagonals ✓
∴ All sides equal (√34) and diagonals equal (√68) → ABCD is a square.
✂️ Concept 2 — Section Formula
Q4. In what ratio does the y-axis divide the segment joining A(−3, 4) and B(1, −2)? Medium▼
1
On the y-axis, x-coordinate = 0. Let ratio = k : 1.
2
Apply x-part of section formula: 0 = (k×1 + 1×(−3))/(k+1)
3
k − 3 = 0 → k = 3
4
Find y-coordinate of the point: y = (3×(−2) + 1×4)/(3+1) = (−6+4)/4 = −2/4 = −1/2
∴ The y-axis divides AB in the ratio 3 : 1. The point is (0, −½).
Q5. Find the coordinates of the point that divides the join of (2, −3) and (5, 6) externally in the ratio 2 : 1. Hard▼
1
External section formula: P = ( (mx₂−nx₁)/(m−n) , (my₂−ny₁)/(m−n) )
2
m=2, n=1, (x₁,y₁)=(2,−3), (x₂,y₂)=(5,6).
3
x = (2×5 − 1×2)/(2−1) = (10−2)/1 = 8
4
y = (2×6 − 1×(−3))/(2−1) = (12+3)/1 = 15
∴ The point is (8, 15).
Q6. Three vertices of a parallelogram are A(1, 2), B(4, 3), C(6, 6). Find the fourth vertex D. Medium▼
1
In a parallelogram, diagonals bisect each other. So midpoint of AC = midpoint of BD.
2
Midpoint of AC = ((1+6)/2, (2+6)/2) = (3.5, 4)
3
Let D = (x, y). Midpoint of BD = ((4+x)/2, (3+y)/2). Set equal to (3.5, 4).
4
(4+x)/2 = 3.5 → x = 3; (3+y)/2 = 4 → y = 5
∴ Fourth vertex D = (3, 5).
📐 Concept 3 — Area & Collinearity
Q7. If the points (2, 3), (k, −1) and (0, 4) are collinear, find k. Medium▼
1
For collinear points, Area of triangle = 0. Use: ½|x₁(y₂−y₃)+x₂(y₃−y₁)+x₃(y₁−y₂)| = 0
2
Substitute: (x₁,y₁)=(2,3), (x₂,y₂)=(k,−1), (x₃,y₃)=(0,4)
3
2(−1−4) + k(4−3) + 0(3−(−1)) = 0
4
2(−5) + k(1) + 0 = 0 → −10 + k = 0 → k = 10
∴ k = 10.
Q8. Find the area of the quadrilateral whose vertices, taken in order, are A(−4, 2), B(3, −5), C(3, −2), D(−4, 3). Hard▼
1
Divide quadrilateral ABCD into two triangles by diagonal AC: △ABC and △ACD.
2
Area △ABC: ½|−4(−5−(−2)) + 3(−2−2) + 3(2−(−5))|
3
= ½|−4(−3)+3(−4)+3(7)| = ½|12−12+21| = ½×21 = 10.5
4
Area △ACD: ½|−4(−2−3)+3(3−2)+(−4)(2−(−2))|
5
= ½|−4(−5)+3(1)+(−4)(4)| = ½|20+3−16| = ½×7 = 3.5
6
Total area = 10.5 + 3.5 = 14 sq. units.
∴ Area of quadrilateral ABCD = 14 square units.
🔺 Concept 4 — Centroid & Triangle Properties
Q9. The centroid of a triangle is (1, 4). Two of its vertices are (−1, 6) and (5, 4). Find the third vertex. Easy▼
1
Centroid formula: G = ((x₁+x₂+x₃)/3, (y₁+y₂+y₃)/3)
2
For x: 1 = (−1+5+x₃)/3 → 3 = 4+x₃ → x₃ = −1
3
For y: 4 = (6+4+y₃)/3 → 12 = 10+y₃ → y₃ = 2
∴ Third vertex = (−1, 2).
Q10. Using medians, verify that the centroid of △ with vertices (0,0), (6,0), (0,8) lies at (2, 8/3). Medium▼
1
Direct centroid: G = ((0+6+0)/3, (0+0+8)/3) = (2, 8/3) ✓
2
Midpoint of BC (opposite A): M₁ = ((6+0)/2, (0+8)/2) = (3, 4)
3
Centroid divides median AM₁ in 2:1. Point on AM₁ at ratio 2:1 from A(0,0): x=(2×3+1×0)/3=2, y=(2×4+1×0)/3=8/3 ✓
∴ Centroid = (2, 8/3) verified via both methods.
🔷 Concept 5 — Quadrilateral Classification
Q11. Show that A(3, 2), B(0, 5), C(−3, 2), D(0, −1) form a square. Find its area. Hard▼
1
AB=√[(0−3)²+(5−2)²]=√[9+9]=3√2
2
BC=√[(−3)²+(−3)²]=3√2, CD=3√2, DA=3√2. All sides equal ✓
3
Diagonal AC: AC=√[(3−(−3))²+(2−2)²]=6. Diagonal BD: BD=√[0²+(5−(−1))²]=6. Equal ✓
4
Check diagonals bisect: Midpoint AC = (0, 2); Midpoint BD = (0, 2) ✓
5
Area = side² = (3√2)² = 18 sq. units.
∴ ABCD is a square with area 18 sq. units.
📊 Concept 6 — HOTS / Case Study
Q12. A city map places three roads at A(2, 3), B(10, 3), C(6, 9). A new lamppost is to be placed at the centroid. Another is to be placed on AB such that it divides AB in ratio 3:1. Find both positions and the distance between them. HOTS▼
1
Centroid: G = ((2+10+6)/3, (3+3+9)/3) = (18/3, 15/3) = (6, 5)
2
Point on AB (3:1 internal): m=3, n=1, A=(2,3), B=(10,3).
3
P = ((3×10+1×2)/4, (3×3+1×3)/4) = (32/4, 12/4) = (8, 3)
4
Distance GP: d = √[(8−6)²+(3−5)²] = √[4+4] = √8 = 2√2 ≈ 2.83 units
∴ Centroid lamppost at (6, 5); section lamppost at (8, 3); distance = 2√2 units.
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Section Formula Visualiser

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m = 1 : 1
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Quick Concept Quiz

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ACADEMIA AETERNUM तमसो मा ज्योतिर्गमय · Est. 2025
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Coordinate Geometry Class 10 Notes (NCERT Ch 7) – Easy Guide
Coordinate Geometry Class 10 Notes (NCERT Ch 7) – Easy Guide — Complete Notes & Solutions · academia-aeternum.com
Chapter 7, Coordinate Geometry, equips students with powerful analytical tools to study geometrical figures using numbers. Starting from plotting points on the Cartesian plane, the chapter leads to practical formulas such as the distance formula, midpoint formula, and section formula, which help compute distances, midpoints, and dividing points of line segments. These ideas not only strengthen algebraic skills but also lay the foundation for advanced geometry and real-world applications like…
🎓 Class 10 📐 Mathematics 📖 NCERT ✅ Free Access 🏆 CBSE · JEE
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