Ncert Class 10 Exercise 7.1

NCERT Class 10 Maths Exercise 7.1: A complete walkthrough to Coordinate Geometry

This article provides a complete walkthrough to solving NCERT Class 10 Maths Exercise 7.1, focusing on coordinate geometry. But we'll cover each question in detail, explaining the concepts involved and providing step-by-step solutions. Understanding coordinate geometry is crucial for further studies in mathematics and related fields, so mastering this exercise is a significant step in your academic journey. This guide will not only help you solve the problems but also deepen your understanding of the underlying principles.

Introduction to Coordinate Geometry

Coordinate geometry, also known as analytic geometry, is the study of geometry using a coordinate system. The most common coordinate system is the Cartesian coordinate system, which uses two perpendicular lines, the x-axis and the y-axis, to define a plane. Consider this: essentially, it bridges the gap between algebra and geometry, allowing us to represent geometric shapes and figures using algebraic equations and vice-versa. Every point on this plane is uniquely identified by an ordered pair of coordinates (x, y), representing its horizontal and vertical distance from the origin (0, 0).

Exercise 7.Practically speaking, 1 in your NCERT Class 10 Maths textbook focuses on the application of this coordinate system to various geometric problems, primarily involving distance between two points, section formula, and area of a triangle. Let's dive into the problems one by one.

Exercise 7.1: Problem-by-Problem Solutions

Problem 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)

Solution:

The distance between two points (x₁, y₁) and (x₂, y₂) is given by the distance formula: √[(x₂ - x₁)² + (y₂ - y₁)²]

(a) Let (x₁, y₁) = (2, 3) and (x₂, y₂) = (4, 1). Distance = √[(4 - 2)² + (1 - 3)²] = √[2² + (-2)²] = √(4 + 4) = √8 = 2√2

(b) Let (x₁, y₁) = (-5, 7) and (x₂, y₂) = (-1, 3). Distance = √[(-1 - (-5))² + (3 - 7)²] = √[4² + (-4)²] = √(16 + 16) = √32 = 4√2

(c) Let (x₁, y₁) = (a, b) and (x₂, y₂) = (-a, -b). Distance = √[(-a - a)² + (-b - b)²] = √[(-2a)² + (-2b)²] = √(4a² + 4b²) = 2√(a² + b²)

Problem 2: Find the distance between the points (0, 0) and (36, 15).

Solution:

This is a straightforward application of the distance formula. Let (x₁, y₁) = (0, 0) and (x₂, y₂) = (36, 15).

Distance = √[(36 - 0)² + (15 - 0)²] = √(36² + 15²) = √(1296 + 225) = √1521 = 39

Problem 3: Determine if the points (1, 5), (2, 3), and (-2, -11) are collinear.

Solution:

Points are collinear if they lie on the same straight line. Worth adding: we can check this by calculating the distances between each pair of points. If the sum of two distances is equal to the third distance, the points are collinear.

Let A = (1, 5), B = (2, 3), and C = (-2, -11).

AB = √[(2 - 1)² + (3 - 5)²] = √(1 + 4) = √5 BC = √[(-2 - 2)² + (-11 - 3)²] = √(16 + 196) = √212 AC = √[(-2 - 1)² + (-11 - 5)²] = √(9 + 256) = √265

Since AB + BC ≠ AC, the points are not collinear. Alternatively, we can use the concept of the area of a triangle. If the area of the triangle formed by these three points is zero, then the points are collinear.

Area = 0.5 * |x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂)|

Area = 0.5 * |1(3 - (-11)) + 2(-11 - 5) + (-2)(5 - 3)| = 0.5 * |14 - 32 - 4| = 0.

Since the area is not zero, the points are not collinear.

Problem 4: Check whether (5, -2), (6, 4), and (7, -2) are the vertices of an isosceles triangle.

Solution:

An isosceles triangle has at least two sides of equal length. We calculate the distances between each pair of points:

Let A = (5, -2), B = (6, 4), and C = (7, -2).

AB = √[(6 - 5)² + (4 - (-2))²] = √(1 + 36) = √37 BC = √[(7 - 6)² + (-2 - 4)²] = √(1 + 36) = √37 AC = √[(7 - 5)² + (-2 - (-2))²] = √(4 + 0) = 2

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Since AB = BC = √37, the triangle is an isosceles triangle.

Problem 5: In a classroom, 4 friends are seated at the points A, B, C, and D as shown in the figure. Champa and Chameli walk into the class and after observing for a few minutes Champa asks Chameli, “Don’t you think ABCD is a square?” Chameli disagrees. Using distance formula, find which of them is correct.

(Note: This problem requires a diagram which is not reproducible here. The coordinates of A, B, C, and D need to be provided from the textbook figure to solve this.)

Solution:

To determine if ABCD is a square, we need to verify two conditions:

1. All four sides are equal in length (AB = BC = CD = DA).
2. The diagonals are equal in length (AC = BD) and perpendicular to each other.

Using the distance formula, calculate the lengths of all four sides and both diagonals. But if condition 1 and condition 2 are met, then ABCD is a square. Otherwise, it's not.

Problem 6: Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer:

• (a) (-1, -2), (1, 0), (-1, 2), (-3, 0)
• (b) (-3, 5), (3, 1), (0, 3), (-1, -4)

Solution:

This problem involves calculating the distances between points and potentially the diagonals to determine the type of quadrilateral. Again, we use the distance formula.

(a) Let A = (-1, -2), B = (1, 0), C = (-1, 2), D = (-3, 0). Calculate AB, BC, CD, DA, AC, and BD. Analyze the lengths to determine if it's a parallelogram, rectangle, rhombus, or square.

(b) Follow the same procedure as in (a) for points A = (-3, 5), B = (3, 1), C = (0, 3), D = (-1, -4).

Problem 7: Find the point on the x-axis which is equidistant from (2, -5) and (-2, 9).

Solution:

A point on the x-axis has coordinates (x, 0). We are given two points A(2, -5) and B(-2, 9). Let this point be P(x, 0). The distance PA = PB.

√[(x - 2)² + (0 - (-5))²] = √[(x - (-2))² + (0 - 9)²]

Square both sides and solve for x.

Problem 8: Find the values of y for which the distance between the points P(2, -3) and Q(10, y) is 10 units.

Solution:

We are given the distance PQ = 10. Using the distance formula:

√[(10 - 2)² + (y - (-3))²] = 10

Square both sides and solve for y. You will likely get two solutions for y.

Problem 9: If Q(0, 1) is equidistant from P(5, -3) and R(x, 6), find the values of x.

Solution:

PQ = QR. Use the distance formula to equate the distances PQ and QR, and solve for x.

Problem 10: Find the ratio in which the line segment joining the points (-3, 10) and (6, -8) is divided by (-1, 6).

Solution:

This problem utilizes the section formula. Let the ratio be k:1. The section formula states that the coordinates of a point dividing a line segment with endpoints (x₁, y₁) and (x₂, y₂) in the ratio m:n are given by:

[(mx₂ + nx₁) / (m + n), (my₂ + ny₁) / (m + n)]

In this case, we have (-3, 10) and (6, -8) as the endpoints, and (-1, 6) as the dividing point. Substitute the values into the section formula and solve for k.

Conclusion

This complete walkthrough to NCERT Class 10 Maths Exercise 7.By understanding the concepts of coordinate geometry, the distance formula, the section formula, and the area of a triangle, you can successfully tackle these problems and build a strong foundation for future mathematical studies. 1 provides a detailed explanation and step-by-step solutions for each problem. Remember to practice regularly and thoroughly understand the underlying principles to achieve mastery. Good luck!