Coordinate Geometry: Class X

Coordinate Geometry is an essential chapter in Class X Mathematics, as it lays the foundation for understanding the geometric representation of algebraic equations. This chapter is covered extensively in both RS Aggarwal and RD Sharma textbooks. In this blog article, we will explore the key concepts of Coordinate Geometry and provide solutions to all the questions from RS Aggarwal and RD Sharma.

Key Concepts

**1. Cartesian Coordinate System**

The Cartesian coordinate system is a two-dimensional plane defined by two perpendicular axes: the x-axis (horizontal) and the y-axis (vertical). The point of intersection of these axes is called the origin, denoted by (0, 0).

**2. Coordinates of a Point**

A point in the Cartesian plane is represented by an ordered pair of numbers (x, y), where ‘x’ is the abscissa (horizontal distance from the origin) and ‘y’ is the ordinate (vertical distance from the origin).

**3. Distance Formula**

The distance between two points ((x_1, y_1)) and ((x_2, y_2)) is given by:

d=(x2−x1)/2+(y2−y1)/2**4. Section Formula**

The coordinates of a point that divides the line segment joining two points ((x_1, y_1)) and ((x_2, y_2)) in the ratio (m_1 : m_2) are given by:

(m1+m2m1x2+m2x1,m1+m2m1y2+m2y1)**5. Area of a Triangle**

The area of a triangle formed by three points ((x_1, y_1)), ((x_2, y_2)), and ((x_3, y_3)) is given by:

Area=1/2∣x1(y2−y3)+x2(y3−y1)+x3(y1−y2)∣**RS Aggarwal Solutions****Exercise 6A**

Find the distance between the points (3, 4) and (7, 1).

Solution: Using the distance formula, (d = \sqrt{(7 – 3)^2 + (1 – 4)^2} = \sqrt{16 + 9} = 5).**Find the coordinates of the point which divides the line segment joining (2, -3) and (5, 6) in the ratio 2:3.**

Solution: Using the section formula, the coordinates are (\left( \frac{2 \cdot 5 + 3 \cdot 2}{2 + 3}, \frac{2 \cdot 6 + 3 \cdot (-3)}{2 + 3} \right) = \left( \frac{10 + 6}{5}, \frac{12 – 9}{5} \right) = \left( 3.2, 0.6 \right)).**Exercise 6B**

Find the area of the triangle with vertices (1, 2), (3, 4), and (5, 6).

Solution: Using the area formula, (\text{Area} = \frac{1}{2} \left| 1(4 – 6) + 3(6 – 2) + 5(2 – 4) \right| = \frac{1}{2} \left| -2 + 12 – 10 \right| = 0).**RD Sharma Solutions****Exercise 14.1**

On which axis do the following points lie? (i) P (5, 0) (ii) Q (0, -2) (iii) R (-4, 0) (iv) S (0, 5)

Solution: (i) P (5, 0) lies on the x-axis (ii) Q (0, -2) lies on the y-axis (iii) R (-4, 0) lies on the x-axis (iv) S (0, 5) lies on the y-axis1.**Ji**

Solution: Using the distance formula, (d = \sqrt{(-1 + 6)^2 + (-5 – 7)^2} = \sqrt{25 + 144} = 13).

Distance formula

D =√{x1-x2}^2+{y1-y2}^2

D=√{x2-x1}^2+{y2-y1}^2

Mid-Point Formula

(X1+x2)/2,(y1+y2)/2

(X2+X1),(Y2+Y1)/2

Sectional formula

Internal sectional formula

External sectional formula

Find the value of ‘a’ when the distance between the points (3, a) and (4, 1) is (\sqrt{10}).

Solution: Using the distance formula, (\sqrt{(4 – 3)^2 + (1 – a)^2} = \sqrt{10}). Simplifying, we get (1 + (1 – a)^2 = 10). Therefore, ((1 – a)^2 = 9), giving (a = -2) or (a = 4).

Coordinate Geometry is a fascinating chapter that combines algebra and geometry to solve various problems. By understanding the key concepts and practicing the exercises from RS Aggarwal and RD Sharma, students can master this chapter and excel in their exams.

1: RD Sharma Solutions for Class 10 Maths Chapter 14 Co-ordinate Geometry

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