CBSE Class 10 Maths Revision Notes Chapter 7

Class 10 Mathematics Revision Notes for Coordinate Geometry of Chapter 7

Class 10 Mathematics Chapter 7 Notes are designed for students preparing for the board examinations. Candidates can rely on the information provided in the notes as they are created according to the NCERT Books and CBSE guidelines. Simple language is used in the notes so that one can understand all the concepts clearly. Students can access the Class 10 Chapter 7 Mathematics Notes from the Extramarks website. All the important topics such as distance formula, arrow method, etc. are covered in Chapter 7 Mathematics Class 10 Notes. In addition, students can also access the other study material such as CBSE extra questions, CBSE Sample papers, etc without any hassles. 

Class 10 Mathematics Revision Notes for Coordinate Geometry of Chapter 7 -(Add revision notes PDF)

Access Class 10 Mathematics Chapter 7 – Coordinate Geometry

Introduction

We need two coordinate axes to find a point’s location on a plane. The x-coordinate, also known as the abscissa, is a point’s separation from the y-axis. The y-coordinate, also known as the ordinate, refers to a point’s separation from the x-axis. A point on the x-axis has coordinates of the form (x, 0), and a point on the y-axis has coordinates of the form (0, y).

Students will learn how to calculate the distance between any two points whose coordinates are provided, as well as how to calculate the area of any triangle formed by any three given points, in this chapter. The coordinates of the point that divides a line segment connecting two given points in a specific ratio will also be covered in this lesson.

Important Terms and Concepts:

The important concepts and terms of coordinate geometry are as follows: 

• The perpendicular lines XOX and YOY intersecting at o are the coordinate axes. 
• The plane is divided into four different quadrants by these axes. 
• One can locate the axes on the Cartesian plane. 
• The XOX is known as the x-axis and YOY is called the y-axis. These axes are seen horizontally and vertically on the graph. 
• O is the point of intersection of the axes and is known as the origin. 
• Abscissae are the values of x measured from origin O along the x-axis. The values of OX are positive whereas the values of OX’ are negative. 
• Similarly, the ordinate refers to the values of y measured along the y axis from O. y has positive values for OY and negative ones for OY’. 
• Coordinates of a point are ordered pairs containing abscissas and ordinates. 

Distance Formula:

The distance formula to measure the distance between any two points is (x1, y1), and (x2, y2) is given by : 

d = √[x2 – x1)2+(y2 – y1)2]

Where d = distance between the two points (x1,y1) and (x2,y2).

Section Formula:

If the point P (x, y) divides the line segment joining A (x1, y1) and B(x2, y2) internally in the ratio m:n then the coordinates of P can be given through the section formula as: 

P (x, y) = (mx2 + nx1 / m + n, my2 + ny1 / m + n)

Midpoint Formula:

The midpoint of any particular line segment divides it in a 1:1 ratio. 

One can understand the midpoint formula through the diagram given below: 

The coordinates of the midpoint (P) of the line segment joining the A(x1, y1) and B(x2, y2) are as follows: 

p (x,y) = (x1 + x2 / 2 , y1 + y2 / 2)

Let us consider an example: 

• What is the midpoint of PQ, whose coordinates are P(-3, 3) and Q(1, 4)? 

Solution 

P(-3, 3) and Q(1, 4) are the points of line segment PQ. 

Now using the midpoint formula we get: 

Midpoint of PQ = (-3 + 1 / 2, -3 + 4 / 2)

= (-2 / 2 , 1 / 2)

= (-1, ½)

The centroid of a Triangle:

If A(x1, y1), B(x2, y2), and C(x3, y3) are the three vertices of ΔABC, then the coordinates of centroid P are given by: 

p (x, y) = (x1 + x2 + x3 / 3 , y1 + y2 + y3 / 3)

Example: find the coordinates of the centroid of a triangle whose vertices are (-1, -3), (2, 1), and (8, -4). 

Solution: 

The coordinates are (-1, -3), (2, 1) and (8, -4).

The centroid of the triangle will be: 

G = ((x1+x2+x3)/3 , (y1+y2+y3)/3 )

G = ((-1+2+8)/3 , (-3+1-4)/3)

G = ( 9/3 , -6/3)

G = (3, -2)

The centroid of the triangle is G = (3, -2). 

Area of the Triangle:

To find the area of the triangle if A(x1, y1), B(x2, y2), and C(x3, y3) are vertices of triangle ABC, then the area will be given by 

A = (1/2)[x1(y2 − y3) + x2(y3 − y1) + x3(y1 − y2)]

Where A = area of the triangle 

Example: find the area of triangle ABC whose vertices are A(1, 2), B(4, 2), and C(3, 5).

Solution: 

Using the formula for the area of the triangle, we get: 

A = (1/2) [x1 (y2 – y3 ) + x2 (y3 – y1 ) + x3(y1 – y2)]

A = (1/2) [1(2 – 5) + 4(5 – 2) + 3(2 – 2)]

A = (1/2) [-3 + 12]

So the area will be 9/2 square units. 

Arrow Method:

It is used to find the formula of the area of a triangle. So 

½ [x1 Y1 1]

   [x2 y2 1]

   [x3 y3 1]

½ [x1 (y2 – y3) + x2 (y3 – y1) + x3 (y1 – y2)]

Note: 

• If points A, B, and C are in an anti-clockwise direction, then the area will be positive. 
• If points A, B, and C are in a clockwise direction, then the area will be negative. 
• If the area of the triangle is zero, then three points A, B and C will be collinear.