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.
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.
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.
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.
The Coordinate Geometry chapter in Class 10 Maths offers several benefits:
(Session 2025 - 26)