**Question 1:** **On a graph paper, draw a horizontal line Xâ€™OX and a vertical line YoYâ€™ as the x-axis and the y-axis respectively.**

**2x – 3y – 17 = 0,Â ****4x + y – 13 =0 **

**Solution:**

Given equations are 2x – 3y – 17 = 0 and 4x + y – 13 =0

Graph of 2x – 3y – 17 = 0:

2x – 3y – 17 = 0 \( \Rightarrow y = \frac{2x – 17}{3} \)

Putting x = 1, we get y = -5

Putting x = 4, we get y = -3

Putting x = 7, we get y = -1

Hence, the table is,

x |
1 | 4 | 7 |

y |
-5 | -3 | -1 |

Plot the points A(1, -5), B(4, -3) and C(7, -1) on the graph paper. Join AB and BC to get the graph line AC. Extend it both ways.

Thus, line AC is the graph of 2x – 3y – 17 = 0.

Graph of 4x + y – 13 =0 :

4x + y – 13 =0Â \( \Rightarrow y = -4x + 13 \)

Putting x = 4, we get y = -3

Putting x = 2, we get y = 5

Putting x = 3, we get y = 1

Hence, the table is,

x |
4 |
2 |
3 |

y |
-3 |
5 |
1 |

Now, on the same graph paper as above plot the points P(2, 5) and Q(3, 1). The point B(4, -3) has already been plotted. Join PQ and QB to get PB. The line PB is the graph of the equation 4x + y – 13 =0 is the solution of the given system of equations.

These graph lines intersect the x-axis at R and S. the region bounded by these lines and the x-axis has been shaded.

**Question 2:** **On a graph paper, draw a horizontal line Xâ€™OX and a vertical line YoYâ€™ as the x-axis and the y-axis respectively.**

**4x – y = 4,Â ****3x + 2y = 14 **

**Solution:**

Given equations are 4x – y = 4 and 3x + 2y = 14

Graph of 4x – y = 4:

4x – y = 4 \( \Rightarrow y = 4x – 4 \)

Putting x = 0, we get y = -4

Putting x = 1, we get y = 0

Putting x = 2, we get y = 4

Hence, the table is,

x |
0 |
1 |
2 |

y |
-4 |
0 |
4 |

Plot the points A(0, -4), B(1, 0) and C(2, 4) on the graph paper. Join AB and BC to get the graph line AC. Extend it both ways.

Thus, line AC is the graph of 4x – y = 4.

Graph of 3x + 2y = 14 :

3x + 2y = 14 Â \( \Rightarrow y =\frac{14 – 3x}{2} \)

Putting x = 0, we get y = 7

Putting x = 2, we get y = 4

Putting x = 4, we get y = 1

Hence, the table is,

x |
0 | 2 | 4 |

y |
7 | 4 | 1 |

Now, on the same graph paper as above plot the points P(0, 7) and Q(4, 1). The point C(2, 4) has already been plotted. Join PC and CQ to get PQ.

The line PQ is the graph of the equation 3x + 2y = 14. The two graph lines intersect at point C(2, 4)

Therefore, x = 2, y = 4 is the solution of the given system of equations. The region bounded by these lines and y-axis has been shown by the shaded area.

**Question 3:** **On a graph paper, draw a horizontal line Xâ€™OX and a vertical line YoYâ€™ as the x-axis and the y-axis respectively.**

**2x – y = 1,Â ****x – y = -1 **

**Solution:**

Given equations are 2x – y = 1 and x – y = -1

Graph of 2x – y = 1:

2x – y = 1 \( \Rightarrow y = 2x – 1 \)

Putting x = 1, we get y = 1

Putting x = 2, we get y = 3

Putting x = 0, we get y = -1

Hence, the table is,

x |
1 | 2 | 0 |

y |
1 | 3 | -1 |

Plot the points A(1, 1), B(2, 3) and C(0, -1) on the graph paper. Join AB and AC to get the graph line BC. Extend it both ways.

Thus, line BC is the graph of 2x – y = 1.

Graph of x – y = -1:

x – y = -1 \( \Rightarrow y =x + 1 \)

Putting x = 1, we get y = 2

Putting x = 2, we get y = 3

Putting x = 0, we get y = 1

Hence, the table is:

x |
1 |
2 |
0 |

y |
2 |
3 |
1 |

Now, on the same graph paper as above plot the points P(1, 2) and Q(0, 1). The point B(2, 3) has already been plotted. Join PB and PQ to get BQ.

The line BQ is the graph of the equation x – y = -1. The two graph lines intersect at point B(2, 3)

Therefore, x = 2, y = 3 is the solution of the given system of equations. The region bounded by these lines and y-axis has been shown by the shaded area.

**Question 4:**

**2x – 5y + 4 = 0,**

**2x + y – 8 = 0 **

**Solution:**

Given equations are 2x – 5y + 4 = 0 and 2x + y – 8 = 0

Graph of 2x – 5y + 4 = 0:

2x – 5y + 4 = 0 \( \Rightarrow y = \frac{2x + 4}{5} \)

Putting x = 3, we get y = 2

Putting x = -2, we get y = 0

Putting x = 8, we get y = 4

Hence, the table is,

x |
3 | -2 | 8 |

y |
2 | 0 | 4 |

Plot the points A(3, 2), B(-2, 0) and C(8, 4) on the graph paper. Join AB and AC to get the graph line BC. Extend it both ways.

Thus, line BC is the graph of 2x – 5y + 4 = 0.

Graph of 2x + y – 8 = 0:

2x + y – 8 = 0 \( \Rightarrow y = -2x + 8 \)

Putting x = 1, we get y = 6

Putting x = 2, we get y = 4

Putting x = 3, we get y = 4

Hence, the table is,

x |
1 | 2 | 3 |

y |
6 | 4 | 2 |

Now, on the same graph paper as above plot the points P(1, 6) and Q(2, 4). The point A(3, 2) has already been plotted. Join PA.

Thus, line PA is the graph of 2x + y – 8 = 0. On extending the graph lines on both sides, we find that these graph lines intersect the y-axis at the point R(0, 8) and S(0, 0.8).

**Question 5:**

**4x – 5y – 20 = 0 and 3x + 5y – 15 = 0 **

**Solution:**

Given equations are 4x – 5y – 20 = 0 and 3x + 5y – 15 = 0

Graph of 4x – 5y – 20 = 0:

4x – 5y – 20 = 0 \( \Rightarrow y = \frac{4x – 20}{5} \)

Putting x = 0, we get y = -4

Putting x = 5, we get y = 0

Putting x = 10, we get y = 4

Hence, the table is:

x |
0 | 5 | 10 |

y |
-4 | 0 | 4 |

Plot the points A(0, -4), B(5, 0) and C(10, 4) on the graph paper. Join AB and BC to get the graph line AC. Extend it both ways.

Thus, line AC is the graph of 4x – 5y – 20 = 0.

Graph of 3x + 5y – 15 = 0:

3x + 5y – 15 = 0 \( \Rightarrow y = \frac{-3x + 15}{5} \)

Putting x = 5, we get y = 0

Putting x = 0, we get y = 3

Putting x = -5, we get y = 6

Hence, the table is:

x |
5 | 0 | -5 |

y |
0 | 3 | 6 |

Now, on the same graph paper as above plot the points P(0, 3) and Q(-5, 6). The point B(5, 0) has already been plotted. Join PQ and PB to get the line QB.

Thus, line QB is the graph of 3x + 5y – 15 = 0. The two graph lines intersect at B(5, 0).

Hence, x = 5, y = 0 is the solution of the given system of equations.

Clearly, the vertices of \( \Delta PBA \)

**Question 6:**

**4x – 3y + 4 = 0,**

**4x + 3y – 20 = 0 **

**Solution:**

Given equations are 4x – 3y + 4 = 0 and 4x + 3y – 20 = 0

Graph of 4x – 3y + 4 = 0:

4x – 3y + 4 = 0 \( \Rightarrow y = \frac{4x + 4}{3} \)

Putting x = -1, we get y = 0

Putting x = 2, we get y = 4

Putting x = 5, we get y = 8

Hence, the table is:

x |
-1 | 2 | 5 |

y |
0 | 4 | 8 |

Plot the points A(-1, 0), B(2, 4) and C(5, 8) on the graph paper. Join AB and BC to get the graph line AC. Extend it both ways.

Thus, line AC is the graph of 4x – 3y + 4 = 0.

Graph of 4x + 3y – 20 = 0:

4x + 3y – 20 = 0 \( \Rightarrow y = \frac{-4x + 20}{3} \)

Putting x = 2, we get y = 4

Putting x = -1, we get y = 8

Putting x = 5, we get y = 0

Hence, the table is:

x |
2 | -1 | 5 |

y |
4 | 8 | 0 |

Now, on the same graph paper as above plot the points P(-1, 8) and Q(5, 0). The point B(2, 4) has already been plotted. Join QB and PB to get the line PQ.

Thus, line PQ is the graph of 4x + 3y – 20 = 0. The two graph lines intersect at B(2, 4).

Hence, x = 2, y = 4 is the solution of the given system of equations.

Clearly, the vertices of \( \Delta ABQ \)

Consider the triangle \( \Delta ABQ \)

Height of the triangle = 4 units and base(AQ) = 6 units

Area of the triangle \( \Delta ABQ \)

\( Area = \frac{1}{2} \times Base \times height = \frac{1}{2} \times 4 \times 6 \)

Area of \( \Delta ABQ \)

**Question 7:**

**x – y + 1 = 0,Â ****3x + 2y – 12 = 0 **

**Solution:**

Given equations are x – y + 1 = 0 and 3x + 2y – 12 = 0

Graph of x – y + 1 = 0:

x – y + 1 = 0 \( \Rightarrow y = x + 1 \)

Putting x = -1, we get y = 0

Putting x = 1, we get y = 2

Putting x = 2, we get y = 3

Hence, the table is,

x |
-1 | 1 | 2 |

y |
0 | 2 | 3 |

Plot the points A(-1, 0), B(1, 2) and C(2, 3) on the graph paper. Join AB and BC to get the graph line AC. Extend it both ways.

Thus, line AC is the graph of x – y + 1 = 0.

Graph of 3x + 2y – 12 = 0:

3x + 2y – 12 = 0 \( \Rightarrow y = \frac{-3x + 12}{2} \)

Putting x = 0, we get y = 6

Putting x = 2, we get y = 3

Putting x = 4, we get y = 0

Hence, the table is:

x |
0 | 2 | 4 |

y |
6 | 3 | 0 |

Now, on the same graph paper as above plot the points P(0, 6) and Q(4, 0). The point C(2, 3) has already been plotted. Join PC and CQ to get the line PQ.

Thus, line PQ is the graph of 3x + 2y – 12 = 0. The two graph lines intersect at C(2, 3).

Hence, x = 2, y = 3 is the solution of the given system of equations.

Clearly, the vertices of \( \Delta ACQ \)

Consider the triangle \( \Delta ACQ \)

Height of the triangle = 3 units and base(AQ) = 5 units

Area of the triangle \( \Delta ACQ \)

\( Area = \frac{1}{2} \times Base \times height = \frac{1}{2} \times 3 \times 5 \)

Area of \( \Delta ACQ \)