# Exercise 3.3 Coordinate Geometry NCERT Solutions Class 9

Exercise 3.3

## Chapter 3 Ex.3.3 Question 1

In which quadrant or on which axis do each of the points \((–2, 4), (3, –1), (–1, 0), (1, 2)\) and \((–3, –5)\) lie? Verify your answer by locating them on the Cartesian plane.

**Solution**

**Video Solution**

**Steps:**

To determine the quadrant or axis of the points

\((–2, 4), (3, –1), (–1, 0), (1, 2)\)

and \((–3, –5)\).

Plot the points

\((–2, 4), (3, –1), (–1, 0), (1, 2)\) and

\((–3, –5)\) on the graph, to get

From the figure above, we can conclude that the points

- Point \((–2, 4)\) lie in \(2^\rm{nd}\)
^{ }quadrant^{ } - Point \((3, –1)\) lie in \(4^\rm{th}\)
^{ }quadrant - Point \((–1, 0)\) lie on the negative
*\(x-\)axis*. - Point \((1, 2)\) lie in \(1^\rm{st}\)
^{ }quadrant^{. } - Point \((–3, –5)\) lie in \(3^\rm{rd} \)quadrant
^{. }

## Chapter 3 Ex.3.3 Question 2

Plot the points (\(x, y\)) given in the following table on the plane, choosing suitable units of distance on the axes.

\({x}\) | \(-2\) | \(-1\) | \(0\) | \(1\) | \(3\) |

\(y\) | \(8\) | \(7\) | \(-1.25\) | \(3\) | \(-1\) |

**Solution**

**Video Solution**

**Steps:**

Given,

\(x\) | \(-2\) | \(-1\) | \(0\) | \(1\) | \(3\) |

\(y\) | \(8\) | \(7\) | \(-1.25\) | \(3\) | \(-1\) |

Draw \(x\)-axis and \(y\)-axis as the co-ordinate axes and mark their point of intersection \(O\) as the Origin (\(0, 0\)).

In order to plot the points given in the table provided above,

- Let us mark the first point as \(A\,(–2, 8)\)
- Take \(2\) units on the negative \(x-\)axis and then \(8\) units on positive \(y-\)axis.

- Let us mark the second point as \(B\,(–1, 7)\)
- Take \(-1\) unit on negative \(x-\)axis and then \(7\) units on positive \(y-\)axis.

- Let us mark the third point as \(C\,(0, –1.25),\)
- Take \(1.25\) units below
*\(x-\)axis*on negative \(y-\)axis.

- Take \(1.25\) units below
- Let us mark the fourth point as \(D\,(1, 3)\)
- Take \(1\) unit on positive \(x-\)axis and then \(3\) units on positive \(y-\)axis.

- Let us mark the fifth point as \(E\,(3, –1)\)
- Take \(3\) units on positive \(x-\)axis and then move \(-1\) unit on negative \(x-\)axis.