NCERT Solutions Class 10 Maths Chapter 7 Coordinate Geometry: CBSE Class 10 Mathematics (2025–2026)

In CBSE Class 10 Mathematics, Chapter 7 Coordinate Geometry is one of the most scoring and conceptually rewarding units for the 2025–2026 board exam 10. Coordinate geometry acts as a bridge between algebra and Euclidean geometry, allowing geometric shapes to be analyzed using algebraic equations and ordered coordinate pairs (x, y) on the Cartesian plane. With these comprehensive NCERT Solutions Class 10 Maths Chapter 7 Coordinate Geometry, students can master essential formulas, geometric proofs, and high-frequency examination questions with step-by-step precision.

CBSE Class 10 Coordinate Geometry: Syllabus & Exam Weightage (2025–2026)

In the updated CBSE Mathematics curriculum for Class 10, Coordinate Geometry carries a direct weightage of approximately 6 marks in the annual board examination. The questions are distributed across Multiple Choice Questions (Section A), short-answer questions (Section B/C), and structured Case Study-based questions (Section E).

Under the rationalized curriculum, the chapter focuses on two primary mathematical pillars:

• The Distance Formula: Finding distances between points, testing collinearity, and determining the geometric nature of triangles and quadrilaterals.
• The Section Formula: Calculating coordinates of points dividing a line segment internally in a given ratio, finding midpoints, and determining unknown ratios.

Real-World Application: Coordinate geometry is the mathematical backbone of modern Global Positioning Systems (GPS), air traffic navigation, computer graphics, and robotic path planning, where physical locations are mapped onto multidimensional coordinate grids.

Key Concepts and Core Formulas

1. The Distance Formula

The distance d between any two points P(x₁, y₁) and Q(x₂, y₂) in the Cartesian plane is derived from the Pythagorean theorem:

PQ = √[(x₂ − x₁)² + (y₂ − y₁)²]

Special Cases and Properties:

• Distance from the Origin: The distance of a point P(x, y) from the origin O(0, 0) is given by OP = √(x² + y²).
• Condition for Collinearity: Three points A, B, and C are collinear if and only if the sum of the lengths of any two segments equals the length of the third segment (e.g., AB + BC = AC).
• Geometric Classifications using Distance:
  • Equilateral Triangle: AB = BC = CA
  • Isosceles Triangle: Any two sides are equal.
  • Right-angled Triangle: AB² + BC² = AC² (Converse of Pythagoras Theorem).
  • Square: All 4 sides equal and both diagonals equal (AB = BC = CD = DA and AC = BD).
  • Rhombus: All 4 sides equal, but diagonals are unequal.
  • Rectangle: Opposite sides equal and diagonals equal.
  • Parallelogram: Opposite sides equal, but diagonals are unequal.

2. The Section Formula (Internal Division)

If a point P(x, y) divides the line segment joining A(x₁, y₁) and B(x₂, y₂) internally in the ratio m₁ : m₂, the coordinates of P are:

x = (m₁x₂ + m₂x₁) / (m₁ + m₂)

y = (m₁y₂ + m₂y₁) / (m₁ + m₂)

Pro-Tip for Calculations: When the ratio is unknown, assume the ratio to be k : 1. The coordinates of P then become [(kx2 + x1)/(k + 1), (ky2 + y1)/(k + 1)], which reduces the calculation to a single variable.

3. The Mid-Point Formula and Centroid

When the dividing point P is the exact midpoint of AB, the ratio is 1 : 1 (i.e., m₁ = m₂ = 1):

Midpoint M = [ (x₁ + x₂)/2 , (y₁ + y₂)/2 ]

Centroid of a Triangle: The centroid G of a triangle with vertices (x₁, y₁), (x₂, y₂), and (x₃, y₃) is given by:

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

Important CBSE Class 10 Questions with Detailed Solutions

Question 1 (2 Marks — Distance on Axis)

Problem: Find a point on the y-axis which is equidistant from the points A(6, 5) and B(−4, 3).

Solution:

1. Any point lying on the y-axis has its x-coordinate equal to 0. Let the required point be P(0, y).
2. Since point P is equidistant from A(6, 5) and B(−4, 3), we have PA = PB ⇒ PA² = PB².
3. Using the distance formula:
   PA² = (6 − 0)² + (5 − y)² = 36 + (25 − 10y + y²) = y² − 10y + 61
   PB² = (−4 − 0)² + (3 − y)² = 16 + (9 − 6y + y²) = y² − 6y + 25
4. Equating PA² and PB²:
   y² − 10y + 61 = y² − 6y + 25
   −10y + 6y = 25 − 61
   −4y = −36 ⇒ y = 9

Answer: The required point on the y-axis is (0, 9).

Question 2 (3 Marks — Finding the Ratio of Division)

Problem: In what ratio does the point P(−1, 6) divide the line segment joining the points A(−3, 10) and B(6, −8)?

Solution:

1. Let the point P(−1, 6) divide line segment AB internally in the ratio k : 1.
2. Here, (x₁, y₁) = (−3, 10), (x₂, y₂) = (6, −8), and (x, y) = (−1, 6).
3. By the section formula for the x-coordinate:
   x = (k × x₂ + 1 × x₁) / (k + 1)
   −1 = [k(6) + 1(−3)] / (k + 1)
   −1(k + 1) = 6k − 3
   −k − 1 = 6k − 3
   7k = 2 ⇒ k = 2/7
4. Verification with the y-coordinate:
   y = [ (2/7)(−8) + 10 ] / [ (2/7) + 1 ] = [ −16/7 + 70/7 ] / [ 9/7 ] = (54/7) / (9/7) = 6. (Matches given y-coordinate).

Answer: The point P divides the line segment AB in the ratio 2 : 7.

Question 3 (3 Marks — Vertices of a Parallelogram)

Problem: If A(1, 2), B(4, y), C(x, 6), and D(3, 5) are the vertices of a parallelogram taken in order, find the values of x and y.

Solution:

1. We know that the diagonals of a parallelogram bisect each other. Therefore, the midpoint of diagonal AC is identical to the midpoint of diagonal BD.
2. Midpoint of AC = [ (1 + x)/2 , (2 + 6)/2 ] = [ (1 + x)/2 , 4 ]
3. Midpoint of BD = [ (4 + 3)/2 , (y + 5)/2 ] = [ 7/2 , (y + 5)/2 ]
4. Equating the corresponding x and y coordinates:
   (1 + x)/2 = 7/2 ⇒ 1 + x = 7 ⇒ x = 6
   (y + 5)/2 = 4 ⇒ y + 5 = 8 ⇒ y = 3

Answer: The value of x is 6 and y is 3.

Question 4 (3 Marks — Points of Trisection)

Problem: Find the coordinates of the points of trisection of the line segment joining A(2, −2) and B(−7, 4).

Solution:

1. Points of trisection divide the segment into three equal parts. Let these points be P and Q such that AP = PQ = QB.
2. Point P divides AB internally in the ratio 1 : 2:
   x_P = [ 1(−7) + 2(2) ] / (1 + 2) = (−7 + 4) / 3 = −3 / 3 = −1
   y_P = [ 1(4) + 2(−2) ] / (1 + 2) = (4 − 4) / 3 = 0
   Hence, P = (−1, 0).
3. Point Q is the midpoint of PB (or divides AB in the ratio 2 : 1):
   x_Q = (−1 + (−7)) / 2 = −8 / 2 = −4
   y_Q = (0 + 4) / 2 = 4 / 2 = 2
   Hence, Q = (−4, 2).

Answer: The coordinates of the points of trisection are (−1, 0) and (−4, 2).

Question 5 (4 Marks — CBSE Case Study / Application Question)

Problem: In a sports day event on a rectangular field ABCD, lines are drawn with chalk powder at 1 m intervals. 100 flower pots are placed at a distance of 1 m from each other along AD. Niharika runs 1/4th of distance AD on the 2nd line and plants a green flag. Preet runs 1/5th of distance AD on the 8th line and plants a red flag.

1. Find the coordinates of the positions where the green and red flags are posted.
2. Calculate the shortest distance between both flags.
3. If Rashmi has to post a blue flag exactly halfway between the line segment joining the two flags, at what coordinates should she post it?

Solution:

1. Total distance along AD = 100 m.
   Niharika's position (Green Flag, G): x = 2, y = (1/4) × 100 = 25 ⇒ G(2, 25).
   Preet's position (Red Flag, R): x = 8, y = (1/5) × 100 = 20 ⇒ R(8, 20).
2. Distance between both flags (GR):
   GR = √[ (8 − 2)² + (20 − 25)² ]
   GR = √[ (6)² + (−5)² ] = √[ 36 + 25 ] = √61 meters (≈ 7.81 m).
3. Position of Rashmi's blue flag (Midpoint of GR):
   M = [ (2 + 8)/2 , (25 + 20)/2 ] = [ 10/2 , 45/2 ] = (5, 22.5).
   Rashmi should post her blue flag on the 5th line at a distance of 22.5 meters.

Common Pitfalls and How to Avoid Them

• Negative Sign Errors: When subtracting negative coordinates in the distance formula (e.g., x₂ − (−3) = x₂ + 3), students frequently make sign mistakes. Always place negative values in parentheses before simplifying.
• Confusing Points on Coordinate Axes: Remember that any point on the x-axis is represented as (x, 0), while any point on the y-axis is represented as (0, y).
• Cross-Multiplication in Section Formula: Ensure m₁ multiplies (x₂, y₂) and m₂ multiplies (x₁, y₁). Avoid cross-pairing them with the wrong endpoints.
• Square Root Simplification: Remember that √(a² + b²) ≠ a + b. Compute the sum of squares completely before evaluating the radical.

Effective Preparation Strategy for Board Exam 2025–2026

1. Master Standard Textbook Problems: Solve every problem from NCERT Exercise 7.1 and 7.2 thoroughly, paying close attention to geometric proof problems.
2. Create a Formula Cheat Sheet: Keep a concise reference card with the Distance Formula, Midpoint Formula, and Section Formula for quick revision.
3. Practice Section-Specific Questions: Work through diverse questions ranging from 1-mark MCQs on coordinate quadrants to 4-mark case study problems.
4. Solve Timed Mock Papers: Test your speed and presentation skills under actual exam conditions to ensure accurate geometric justifications.

Where to Practice More

Consistent practice with curriculum-aligned questions is the key to securing a perfect score in CBSE Class 10 Mathematics. Access chapter-wise test series, official CBSE sample papers, and customized question sheets at Theorify QPTool (qptool.theorify.in) to elevate your board exam preparation.