Coordinate Geometry Class 10 Notes

Coordinate Geometry is the branch of mathematics where the relation between geometric shapes and algebraic equations (linear equations and quadratic equations) is studied in terms of a coordinate. In Class 10th Maths Coordinate Geometry is used to understand how geometric concepts relate to algebra. 

1.0The Cartesian Plane

Coordinate Geometry is used to visually represent the position of a given coordinate in a two-dimensional plane, also known as a cartesian plane. The Cartesian plane is a system of two perpendicular axes intersecting each other at 90. The horizontal or x-axis is known as abscissa, and the vertical or y-axis is called the ordinate.  

The abscissa and ordinate intersect at the origin (0,0), dividing the plane into four quadrants, namely: 

• Quadrant I: Both x & y are positive.
• Quadrant II: x is negative, y is positive.
• Quadrant III: Both x & y are negative.
• Quadrant IV: x is positive, y is negative. 

2.0Important Concepts for Class 10 Coordinate Geometry

Coordinates of a Point: 

In coordinate geometry, a point can be represented as an ordered pair (x, y), also known as coordinates, which indicates its position on the Cartesian plane. These coordinates are crucial for understanding how geometric shapes or points are located and measured in the plane.

Distance Formula: 

The distance formula is an important concept of coordinate geometry, used to find straight line distance between two points P(x₁,y₁) & Q(x₂,y₂), as shown in the figure.

The distance(d) may be calculated with the formula, expressed as: 

$d=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}$

This formula for determining the distance between two coordinates is the dominating formula for many coordinate geometry class 10 NCERT solutions.

Section Formula (Internal Division): 

The section formula is used to find the coordinates of a point dividing a given line segment in a certain ratio. If a point, say, P(x, y) divides the line segment joining two points A(x₁, y₁) & B(x₂, y₂) in the ratio m : n, the coordinates of point P are given by:

$x=\frac{n{x}_1+m{x}_2}{m+n},y=\frac{n{y}_1+m{y}_2}{m+n}$

For midpoints where the ratio will be 1:1 then the formula is: 

$x=\frac{x_1+x_2}{2},y=\frac{y_1+y_2}{2}$

The above-mentioned formula is used to solve a wide range of coordinate geometry class 10 important questions, which include finding the ratio dividing the line segments.

Area of a Triangle: 

The area of a triangle which is formed by any three coordinates, say, A(x₁, y₁), B(x₂, y₂), and C(x₃, y₃), in coordinate geometry can be found with the help of the following formula: 

$\text{Area}=\frac{1}{2}|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|$

In coordinate geometry class 10 solutions, you will often come across problems where you need to find the collinearity of any 3 coordinates. The formula mentioned above is used to find the collinearity of the given three points. Collinearly is in the same straight line. The area of the triangle will be 0 if the points are collinear.

Area of triangle = 0; if the points are collinear

3.0Problems On Class 10 Coordinate Geometry

Problem: Find the area of the triangle ABC with A (1, –4) and the mid-points of sides through A being (2, – 1) and (0, – 1).

Solution: Let the other coordinates of the triangle be B (a,b) and C(x,y). Let the midpoints of the triangle be D (2, – 1) and E(0, – 1)

In the side AB, 

$2=\frac{1+a}{2}$, $–1=\frac{-4+b}{2}$

$4=1+a$ $–2=–4+b$

$a=3$   $b=2$

Coordinates of B (3,-3)

In the side AC, 

$0=\frac{1+x}{2}$, $–1=\frac{-4+b}{2}$

$0=1+a$ $–2=–4+b$

$a=–1$   $b=2$

Coordinates of C (-1,2) 

$\text{Area of triangle ABC}=\frac{1}{2}[1(2-2)+3(2+4)–1(-4-2)]$

$=\frac{1}{2}[18+6]$

= 12 square units. 

Problem 2: The point which divides the line segment joining the points (7, –6) and (3, 4) in ratio 1 : 2 internally lies in which section. 

Solution: Coordinates the point dividing the line segment (7, –6) and (3, 4). Let n = 1 and m = 2

$x=\frac{n{x}_1+m{x}_2}{m+n},y=\frac{n{y}_1+m{y}_2}{m+n}$

$x=\frac{2(7)+1(3)}{3},y=\frac{1(-6)+2(4)}{3}$

$x=17/3,y=2/3$

Both the points are positive. Hence, the coordinates lie in the I quadrant. 

Problem 3: What will be the distance between points A (0, 6) and B (0, –2)? 

Solution: Using the distance formula: 

$d=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}$

$d=\sqrt{(0-0{)}^2+(6-(-2){)}^2}$

$d=\sqrt{8^2}=8units$

These were some of the most important questions from the Coordinate Geometry for Board or any other competitive exams. To give your best shot at the board exams this year, you need to solve the previous year's questions pdf.