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NCERT Solutions for Class 10 Maths Chapter 7 - Coordinate Geometry
NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry discusses the numerous aspects of coordinate geometry. It helps us study geometry using algebra and understand algebra with the help of geometry, making it widely applicable in fields such as physics, engineering, navigation, and art.
This article will provide students with high-quality NCERT Solutions for class 10 Maths Chapter 7 Coordinate Geometry exercises devised specifically to help students develop visualization skills, thus enabling them to analyze graphs precisely. Coordinate Geometry class 7 NCERT solutions are developed by ALLEN's subject experts and include the entire chapter concepts as per the latest CBSE curriculum.
1.0Download NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry : Free PDF
Chapter 7 NCERT Solutions of Class 10 Maths is crucial because it teaches students how to calculate the area of the triangle formed by three given points and the distance between two points whose coordinates are given. The foundation for further study of graph-related subjects in upper grades is laid out in NCERT Solutions Chapter 7, and some of these topics are covered in the exercises below.
2.0Exercise-wise NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry
3.0Important Topics - Class 10 Maths Chapter 7 Coordinate Geometry Solutions
Chapter 7, Coordinate Geometry, introduces students to the coordinate plane where points are represented by ordered pairs (x, y). Key concepts include the distance formula to find the distance between two points, the midpoint formula to determine the center point of a line segment, and the section formula for dividing a line segment in a given ratio. These concepts help solve geometric problems algebraically and are essential for understanding the relationship between geometry and algebra.
- The Coordinate Plane (Cartesian Plane)
- Distance Formula
- Section Formula
- Area of Triangles
- Collinearity of 3 Points
4.0General Outline for Class 10 Maths Chapter 7: Coordinate Geometry
Understanding Key Concepts:
- Coordinate Plane: The coordinate plane consists of two axes, X-axis (horizontal) and Y-axis (vertical), forming the origin (0, 0).
- Coordinates of a Point: The position of any point in the plane is represented as (x, y), where:
- x is the horizontal distance from the origin.
- y is the vertical distance from the origin.
- Distance Formula: Used to find the distance between two points and .
- Midpoint Formula: Used to find the midpoint of a line segment joining two points and
- Section Formula: Used to find the coordinates of a point dividing a line segment in a given ratio. Where m:n is the ratio in which P divides AB.
Important Theorems and Formulas:
- Distance between Two Points: Already discussed with the formula.
- Midpoint of a Line Segment: Formula for calculating the midpoint.
- Area of Triangle: Given vertices the area of the triangle is:
- Collinearity of Points: Three points are collinear if the area of the triangle formed by them is zero. You can use the area formula to check this.
Steps for Solving Coordinate Geometry Problems:
- Step 1: Read the problem carefully and identify the known coordinates and unknowns.
- Step 2: Use the appropriate formula depending on what’s being asked:
- Use the distance formula to find the distance between two points.
- Use the midpoint formula to find the midpoint.
- Use the section formula for dividing a line segment in a specific ratio.
- Use the area formula for calculating the area of a triangle.
- Step 3: Substitute the known values into the formula and simplify.
- Step 4: If needed, use algebraic methods to solve for unknown values (e.g., solve for coordinates, distances, or ratios).
- Step 5: Check the solution for consistency and accuracy.
5.0Sample Solutions of NCERT Class 10 Maths Chapter 7 Coordinate Geometry
1. Find the distance between the following pairs of points:
(a) (2, 3), (4, 1)
(b) (-5, 7), (-1, 3)
(c) (a, b), (-a, -b)
Sol.
(a) The given points are: A(2, 3), B(4, 1). Required distance:
AB = BA = √((x₂ - x₁)² + (y₂ - y₁)²)
AB = √((4 - 2)² + (1 - 3)²) = √(2² + (-2)²)
= √(4 + 4) = √8 = 2√2 units
(b) Here x₁ = -5, y₁ = 7 and x₂ = -1, y₂ = 3
Therefore, the required distance:
= √((x₂ - x₁)² + (y₂ - y₁)²)
= √((-1 - (-5))² + (3 - 7)²)
= √((-1 + 5)² + (-4)²)
= √(16 + 16) = √32 = √(2 × 16)
= 4√2 units
(c) Here x₁ = a, y₁ = b and x₂ = -a, y₂ = -b
Therefore, the required distance:
= √((x₂ - x₁)² + (y₂ - y₁)²)
= √((-a - a)² + (-b - b)²)
= √((-2a)² + (-2b)²) = √(4a² + 4b²)
= √(4(a² + b²)) = 2√(a² + b²) units
2. Find the distance between the points (0, 0) and (36, 15).
Sol. Let the points be A(0, 0) and B(36, 15)
Therefore, AB = √((36 - 0)² + (15 - 0)²)
= √(36² + 15²) = √(1296 + 225)
= √1521 = √39² = 39
3. Determine if the points (1, 5), (2, 3) and (-2, -11) are collinear.
Sol. The given points are:
A(1, 5), B(2, 3) and C(-2, -11).
Let us calculate the distance: AB, BC and CA by using the distance formula.
AB = √((2 - 1)² + (3 - 5)²) = √(1² + (-2)²)
= √(1 + 4) = √5 units
BC = √((-2 - 2)² + (-11 - 3)²)
= √((-4)² + (-14)²) = √(16 + 196) = √212
= 2√53 units
CA = √((-2 - 1)² + (-11 - 5)²)
= √((-3)² + (-16)²) = √(9 + 256) = √265
= √(5 × 53) units
From the above we see that: AB + BC ≠ CA
Hence the above stated points A(1, 5), B(2, 3) and C(-2, -11) are not collinear.
4. Find the coordinates of the points of trisection of the line segment joining (4, 1) and (-2, -3).
Sol.
Points P and Q trisect the line segment joining the points A(4, -1) and B(-2, -3), i.e., AP = PQ = QB.
Here, P divides AB in the ratio 1:2 and Q divides AB in the ratio 2:1.
x-coordinate of P = (1 × (-2) + 2 × (4)) / (1 + 2) = 6 / 3 = 2;
y-coordinate of P = (1 × (-3) + 2 × (-1)) / (1 + 2) = -5 / 3
Thus, the coordinates of P are (2, -5/3).
Now, x-coordinate of Q = (2 × (-2) + 1 × (4)) / (2 + 1) = 0;
y-coordinate of Q = (2 × (-3) + 1 × (-1)) / (2 + 1) = -7 / 3
Thus, the coordinates of Q are (0, -7/3).
Hence, the points of trisection are P(2, -5/3) and Q(0, -7/3).
5. Find the area of the triangle whose vertices are :
(i) (2, 3), (-1, 0), (2, -4)
(ii) (-5, -1), (3, -5), (5, 2)
Sol.
(i) Let the vertices of the triangles be A(2, 3), B(-1, 0) and C(2, -4)
Here x₁ = 2, y₁ = 3,
x₂ = -1, y₂ = 0
x₃ = 2, y₃ = -4
Area of a Δ = (1/2) |x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂)|
Area of a Δ = (1/2) |2{0 - (-4)} + (-1){-4 - (3)} + 2{3 - 0}|
= (1/2) |2(0 + 4) + (-1)(-4 - 3) + 2(3)|
= (1/2) |8 + 7 + 6| = (1/2) |21| = 21/2 sq. units
(ii) A(-5, -1), B(3, -5), C(5, 2) are the vertices of the given triangle.
x₁ = -5, x₂ = 3, x₃ = 5; y₁ = -1, y₂ = -5, y₃ = 2.
Area of the ΔABC = (1/2) |x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂)|
= (1/2) |-5 * (-5 - 2) + 3 * (2 + 1) + 5 * (-1 + 5)|
= (1/2) |35 + 9 + 20| = (1/2) |64| = 32 sq. units.
6.0Benefits of Class 10 Maths Chapter 7 Coordinate Geometry
The Coordinate Geometry chapter in Class 10 Maths offers several benefits:
- Foundation for Advanced Mathematics: It introduces the concept of plotting points on a plane, which is essential for higher-level studies in algebra, geometry, and calculus.
- Real-Life Applications: It is used in fields like navigation, computer graphics, engineering, and architecture to represent and analyze positions and movements.
- Problem-Solving Skills: Students develop the ability to solve geometric problems algebraically, improving their analytical thinking.
- Graphing Techniques: Helps students understand how equations and geometric shapes relate to graphs, useful for visualizing data and solving equations.
- Preparation for Future Studies: Coordinate geometry is foundational for studying more advanced topics in mathematics and science, including physics.