## RS Aggarwal Class 9 Solutions Chapter 6

**QUESTION 1:** **Write down the coordinates of each of the points A, B, C, D, E and F shown below draw :**

**SOLUTION:**

Draw the perpendiculars from the AF, BG, CH, DI, and EJ on the x-axis:

**1)** The distance of A from the x-axis= OF= -6 units

The distance of A from y-axis =AF =5 units

**Hence, the coordinate of A is (-6, 5)**

**2)** The distance of B from the x-axis= OG= 5 units

The distance of B from y-axis = BG = 4 units

**Hence, the coordinate of B is (5, 4)**

**3)** The distance of C from the x-axis= OH= -3 units

The distance of C from y-axis = HC =2 units

**Hence, the coordinate of C is (-3, 2)**

**4)** The distance of D from the x-axis= OI= 2 units

The distance of D from y-axis = ID = -2 units

**Hence, the coordinate of D is (-3, -2)**

**5)** The distance of C from the x-axis= OJ= -1 units

The distance of C from y-axis = JE = -4 units

Hence, the coordinate of A is (-1, -4)

Thus, the coordinates of A, B, C, D, and E are respectively:

A (-6, 5), B (5, 4) , C (-3, 2) and E (-1, -4)

**QUESTION 2 :** **Draw the lines X’OX and Y’OY as the coordinate Axes on a paper and plot the following points on it.**

i. P (7, 4)

ii. Q (-5, 3)

iii. R (-6, -3)

iv. S (3, -7)

v. A (6, 0)

vi. B (0, 9)

vii. O (0, 0)

vii. C (-3, -3)

**SOLUTION:**

Let X’OX and Y’OY be the co-ordinate axes.

Fix the side of the small squares as one units:

**i.** Starting from O, take +7 units on the x-axis and then +4 units on the y-axis to obtain the point P (7, 4).

**ii.** Starting from O, take -5 units on the x-axis and then +3 units on the y-axis to obtain the point P (-5, 3).

**iii.** Starting from O, take -6 units on the x-axis and then -3 units on the y-axis to obtain the point P (-6, -3).

**iv.** Starting from O, take 3 units on the x-axis and then -7 units on the y-axis to obtain the point P (3, -7).

**v.** Starting from O, take 6 units on the x-axis and then 0 units on the y-axis to obtain the point P (6, 0).

**vi.** Starting from O, take 0 units on the x-axis and then 9 units on the y-axis to obtain the point P (0, 9).

**vii.** Starting from O, take 0 units on the x-axis and then 0 units on the y-axis to obtain the point P (0, 0).

**viii.** Starting from O, take -3 units on the x-axis and then -3 units on the y-axis to obtain the point P (-3, -3).

**QUESTION 3 :** **On which axes did the following points lie?**

**1. (7, 0)**

**2. (0, -5)**

**3. (0, 1)**

**4. (-4, 0)**

**SOLUTION:**

**1.** In (7, 0) we have the ordinate = 0

Therefore, (7, 0) lies on the x-axis

**2.** In (0, -5), we have the abscissa = 0

Therefore, (0, -5) lies on the y-axis

**3.** In (0, 1), we have the abscissa = 0

Therefore, (0, 1) lies on the y-axis

**4.** In (-4, 0), we have the ordinate = 0

Therefore, (-4, 0) lies on the x-axis

**QUESTION 4 :** **In which quadrant do the given points lie?**

**1. (-6, 5)**

**2. (-3, -2)**

**3. (2, -9)**

**SOLUTION:**

**1.** Point of the type (-, +) lies in the 2^{nd }quadrant.

Therefore, the point (-6, 5) lies in the II quadrant.

**2.** Point of the type (-, -) lie in the 3^{rd }quadrant.

Therefore, the point (-3, -2) lies in the III quadrant.

**3.** Point of the type (+, -) lies in the 4^{th }quadrant.

Therefore, the point (2, -9) lies in the IV quadrant.

**QUESTION 5 :** **Draw the graph of the equation, y = x + 1 **

**SOLUTION:**

The given equation is y = x + 1

Putting x =1, we get y = 1 + 1 = 2

Putting x =2, we get y = 2 + 1 = 3

Thus, we have the following table:

x |
1 | 2 |

Y |
2 | 3 |

On a graph paper, draw the lines X’OX and Y’OY as the x-axis and y-axis respectively.

Now, plot points P (1, 2) and Q (2, 3) on the graph paper.

Join PQ and extend it in both the directions.

Then, line PQ is the graph of the equation y = x + 1.

**QUESTION 6 :** **Draw the graph of the equation, y = 3x + 2 **

**SOLUTION:**

The given equation is y = 3x + 2

Putting x = 1, we get y = (3 * 1) + 2 = 5

Putting x = 2, we get y = (3 * 2) + 2 = 8

Thus, we have the following table:

X |
1 | 2 |

Y |
5 | 8 |

On a graph paper draw the lines X’OX and Y’OY as the x-axis and y-axis respectively.

Now, plot points P (1, 5) and (2, 8).

Join PQ and extend it in both the directions.

Then, line PQ is the graph of the equation y = 3x + 2.

**QUESTION 7 :** **Draw the graph of the equation, y = 5x – 3 **

**SOLUTION:**

The given equation is y = 5x – 3

Putting x = 0, we get y = (5 * 0) – 3 = – 3

Putting x = 1, we get y = (5 * 1) -3 = 2

Thus, we have the following table:

X |
0 | 1 |

Y |
-3 | 2 |

On a graph paper draw the lines X’OX and Y’OY as the x-axis and y-axis respectively.

Now, plot points P (0, -3) and (1, 2).

Join PQ and extend it in both the directions.

Then, line PQ is the graph of the equation y = 5x – 3

**QUESTION 8 :** **Draw the graph of the equation, y = 3x **

**SOLUTION:**

The given equation is y = 3x

Putting x =1, we get y = (3, 1) = 3

Putting x =2, we get y = (3, 2) = 6

Thus, we have the following table:

X |
1 | 2 |

Y |
3 | 6 |

On a graph paper draw the lines X’OX and Y’OY as the x-axis and y-axis respectively.

Now, plot points P (1, 3) and (2, 6)

Join PQ and extend it in both the directions.

Then, line PQ is the graph of the equation y = 3x.

**QUESTION 9 :** **Draw the graph of the equation, y = – x **

**SOLUTION:**

The given equation is y = 3x

Putting x =1, we get y = (3 * 1) = 3

Putting x =2, we get y = (3 * 2) = 6

Thus, we have the following table:

X |
1 | 2 |

Y |
-1 | -2 |

On a graph paper draw the lines X’OX and Y’OY as the x-axis and y-axis respectively.

Now, plot points P (1, -1) and (2, -2).

Join PQ and extend it in both the directions.

Then, line PQ is the graph of the equation y = -x

Related Links |
||
---|---|---|

NCERT Books | NCERT Solutions | RS Aggarwal |

Lakhmir Singh | RD Sharma Solutions | NCERT Solutions Class 6 to 12 |