Chapter 4

Co-ordinate Geometry

Chapter 3 of class 9 is Coordinate Geometry.

A point on a plane, known usually as a Cartesian plane or coordinate plane, can be represented through an ordered pair of real numbers. The unit of mathematics that deciphers geometrical problems using the coordinate system is called coordinate geometry.

Lessons in this chapter are –

  • Introduction to coordinate geometry
  • Explanation of Cartesian System
  • Plotting a point on a plane based on coordinates given

Exercises of NCERT Solutions for Class 9 Chapter 3 Coordinate Geometry

Exercise 1- You need to -

1. Define the position of the table lamp on a study table to another individual

2. Name the horizontal and vertical lines sketched to locate the position of any point on a Cartesian plane

Mark points (x,y) on the plane, appropriate units of distance on the axes

x -2 -1 0 1 3
y 8 7 -1.25 3 -1

 

Answers to NCERT Solution Exercises

1. A perpendicular and a horizontal line are imagined to describe the position of a table lamp on a study table. Suppose a table is a plane with perpendicular line y-axis and horizontal line x-axis. Imagine one point of the table as the origin point of both the x and y axes and the point where the lines intersect.

We have taken y as the length of the table and x as the breadth of the table. The stretch of the source point from both x and y is to be penned in aspects of coordinates. The distance between the point and the x-axis and y-axis, respectively, are X and Y, and as a result, the situation of the lamp in (x, y) coordinate.

2. The horizontal and vertical lines marked on a Cartesian plane are depicted as x and y.

3. The points that would be plotted on (x, y) are –

(-2, 8), (-1, 7), (0, -1.25), (1, 3), (3, -1)

Drawing a graph with the x-axis and y-axis meeting at point O. If 1 unit is equivalent to 1 cm.

For (-2, 8): the fictional lines meet at, from source O, 2 units to the left, and 8 units above.

For (-1, 7): the intersection point of the imaginary lines commences 1 unit to the left of origin O and 7 units above the origin O.

(0, -1.25) on the x-axis 1.25 units to the left of origin O.

(1, 3): I- Quadrant, the intersection point of the imaginary lines that initiates from 1 unit to the right of origin O and three units above origin O.

(3, -1) from the sources O, 3 units right and 1 unit below.

 Q2) A city has two main roads that cross each other at the city's centre. These roads are in the North-South direction and East-West direction. But, every other lane of the city heads parallel to these two roads and are 200 m apart. So, you need to imagine five streets in every direction.

Assume 1 cm = 200 m; make a city model in your notebook or on your desktop. Illustrate the roads by separate lines.

There are various cross-streets in your model. Two streets make one cross-street, one heading in the North-South direction and the other one in the East-West direction.

Every cross lane is referred to in the subsequent setup: If the 2nd street heading in the North-South direction and 5th in the East-West direction meet at any unexpected crossing, we will call this cross-street (2, 5).

By using this design, you need to find out:

a) Number of cross-streets which can be referred to as (4,3)

b) Number of cross-streets which can be referred to as (3, 4).

Solution

a) There is only one lane which you can refer to as (4, 3).

b) There is only one lane which you can refer to as (3, 4).

 

You need to find the explanations to the subsequent questions-

a) Find out the name of horizontal and vertical lines sketched to get the details about the position of any point in the Cartesian plane.

b) You also need to discover the name of every part of the plane created by these two lines.

c) Also, name the point where the two lines will bisect.

Solution-

a) The answer is that the horizontal line is named the x-axis, and the vertical line is the y-axis.

b) The name of every part of the plane formed is called "Quadrant."

c) Points where these two lines cross is called 'Origin.'

Q2) - Check the figure below and articulate.

a.     Coordinates of B

b.    Coordinates of C

c.     The point identified by the coordinates (-3,-5)

d.    The point identified by the coordinates (2,-4)

e.     The abscissa of the point D

f.      The ordinate of the point H

g.     The coordinates of the point

h.    The coordinates of the point M

Solution

After looking at the figure, the solution should be-

a.     Coordinates of B are (-5,2)

b.    Coordinates of C are (5, -5)

c.     The point E is identified by the coordinates (-3,-5)

d.    The point G is identified by the coordinates (2,-4)

e.     The abscissa of the point D is 6

f.      The ordinate of the point H is -3

g.     The coordinates of the point L are (0,5)

h.    The coordinates of the point M are (-3,0)

You need to locate on which axis or in which quadrant are all these points (-2, 4), (3, -1), (-1, 0), (1, 2), and (-3, -5) positioned. It would be best to confirm your solution by finding them on the Cartesian plane.

Solution

The point (-2, 4) is having negative abscissa and positive ordinate.

\( \therefore \) (-2, 4) lies in the 2nd quadrant.

 The point (3, -1) has a positive abscissa and negative ordinate.

(3, -1) lies in the 4th quadrant.

The point (-1, 0) has negative abscissa and zero ordinate.

The point (-1, 0) lies on the negative x-axis.

Point (1, 2) has the abscissa and ordinate positive.

This is because Point (1, 2) lies in the 1st quadrant.

Point (-3, -5) has the abscissa and ordinate negative.

Point (-3, -5) lies in the 3rd quadrant.

These points are plotted in the Cartesian plane as shown in the following figure as A (-2, 4); B(3, -1); C(-l, 0); D(l, 2) and E (-3, -5).

Q2) You need to mark the points (x, y) given in the subsequent table on the plane, picking appropriate distance units on the axes.

Solution

The given points are (-2, 8), (-1, 7), (0, -1.25), (1,3) and (3, -1).

To plot these points-

  • First thing you need to do is, draw X'OX and YOY' as axes.
  • Once this is done, select reasonable units of distance on the axes.
  • To plot (-2, 8), you should begin from O, take (-2) units on the x-axis and then (+8) units on the y-axis. Plot the point as A (-2, 8).
  • To plot (-1, 7), begin from O, take (-1) units on the x-axis and then (+7) units on the y-axis. After this, you need to mark the point as B(-1, 7).
  • To plot (0, -1.25), move along 1.25 units below the x-axis on the y-axis and sketch the point as C(0, -1.25)
  • To plot (1, 3), take (+1) units on the x-axis and then (+3) units on the y-axis. Mark the point as D(1, 3).

(vii) To plot (3, -1), we take (+3) units on the x-axis and then (-1) units on the y-axis. Now, mark the point E(3, -1).

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