Students can track their progress and improvement through regular use of ICSE S Chand Maths Class 9 Solutions Chapter 20 Coordinates and Graphs of Simultaneous Linear Equations Ex 20(B)

## S Chand Class 9 ICSE Maths Solutions Chapter 20 Coordinates and Graphs of Simultaneous Linear Equations Ex 20(B)

Question 1.

Fill in the blanks :

(i) The graph of x = 1 is a line parallel to the ………. axis.

(ii) The graph of y = 1 is a line parallel to the ………. axis.

(iii) The equation ax + by + c = 0 (where a and b are not both zero) is called ……….. equation.

(iv) The graph of 2x = 1 is a line parallel to the ………. axis.

Solution:

(i) The graph of x = 1 is a line parallel to the y-axis.

(ii) The graph of y = 1 is a line parallel to the x- axis.

(iii) The equation ax + by + c = 0 (where a and b are not both zero) is called linear equation.

(iv) The graph of 2x = 1 is a line parallel to the y- axis.

Question 2.

Graph the following equations :

(i) x = 3

Solution:

As x = 3 is a line parallel is y-axis at a distance of 3 units

Now the equation x = 3 can be written as x + 0 y = 3

Giving any values of y as 1,2, 3, etc. we get x = 3

Now plot the points (3, 1),(3, 2),(3, 3), ………… on the graph and join them to get the required line

(ii) y = -4

Solution:

As y = -4 is a line parallel to x-axis at a distance of -4 units. We can write this equation as 0x + y = 4

Now giving any values to x say 1,2,-1 we get y = -4

Now plot the points is (1,-4),(2,-4)(-1, -4). On the graph and join than to get the required line.

(iii) 2x = -7

Solution:

As 2 x = -7 ⇒ x =\(\frac{-7}{2}\) is a line parallel y-axis at a distance of \(\frac{-7}{2}\). We can write it as 2x +0y = -7

Now giving some values to 1,2,3 we get 2x = -7

Now plot the points \(\), \(\left(\frac{-7}{2}, 1\right)\), \(\left(\frac{-7}{2}, 3\right)\) on the graph and join them to get the required line.

(iv) y = 2 x

Solution:

y = 2 x

By giving some value to x we get the corresponding values of y as shown in the following table

X | 1 | 2 | -1 |

y | 2 | 4 | -2 |

Now plot the points (1, 2),(2, 4) and (-1, -2) on the graph and join them to get the required line.

(v) y = -3x

Solution:

y = -3x

By giving some values to x, we get the corresponding values of y as given below :

X | 1 | 0 | -1 |

y | -3 | 0 | 3 |

Now plot the points (1,-3),(0,0) and (-1, 3 ) on the graph and join them to get the required line

(vi) y = x + 1

Solution:

y = x + 1

By giving some values to x we get the corresponding values of y as shown below :

X | 1 | 0 | -1 |

y | 2 | 1 | 0 |

Now plot the points (1,2),(0,1) and (-1, 0 ) on the graph and join them to get the required line

(vii) 2x + y = 14

Solution:

2x + y = 14

⇒ y = 14 – 2x

Giving some different values to x, we get corresponding values of y as given below :

X | 5 | 6 | 7 |

y | 4 | 2 | 0 |

Now plot the points (5, 4),(6, 2) and (7, 0) on the graph and join them to get the required line

(viii) 4x + 3y = 6

Solution:

4x + 3y = 6

⇒ 4x = 6 – 3y

⇒ x = \(\frac{6-3 y}{4}\)

Giving some different values to y we get corresponding values of x as given below :

X | 0 | -3 | 3 |

y | 2 | 6 | -2 |

Now plot the points (0,2),(-3,6) and (3, -2 ) on the graph and join them to get the required line

(ix) x = 3y + 1

Solution:

x = 3y + 1

Giving some different values to y, we get corresponding values of x as given below :

X | 1 | 4 | -2 |

y | 0 | 1 | -1 |

Now plot the points (1, 0),(4, 1) and (-2, -1) on the graph and join them to get the required line.

Question 3.

Draw the graph of equation y = 3x – 4. Find graphically

(i) the values of y, when x = -1.

(ii) the value of x when y = 5.

Solution:

y = 3x – 4

Giving some suitable value to x, we get the corresponding values of y as shown below:

X | 0 | 1 | 2 |

y | -4 | -1 | 2 |

Now plot the points (0, -4),(1, -1) and (2, 2) on the graph and join than to get the required line

(i) From x = -1, draw a perpendicular which meet the line at P. From P draw a line || to x- axis meeting y-axis at -7

∴ If x = -1, then y = -7

(ii) Similarly from x = 5, draw a perpendicular on y-axis which meets the line at Q From Q, draw perpendicular on x-axis meeting at 3

∴ x = 3, y = 5

Question 4.

Find the coordinates of the point where the following lines cut the y-axis.

(i) y = 5x + 1,

(ii) y = 3x – 7

(iii) y = x + 5,

(iv) 3y = 2x + 9.

Solution:

We know that a line cuts the y-axis when x = 0

(i) ∴ y = 5 x + 1 cut the y-axis, when x = 0, When x = 0, then y = 5 × 0 + 1 = 0 + 1 = 1

∴ y = 1

∴ It will cut y-axis at (0, 1)

(ii) y = 3x – 7 will cut the y-axis when x = 0

Now, if x = 0, then y = 3 × 0 – 7 = 0 – 7 = -7

∴ It will cut y-axis at (0, -7)

(iii) y = x + 5 will cut y-axis when x = 0,

Now if x = 0, then y = 0 + 5 = 5

∴ It will cut y-axis at (0, 5)

(iv) 3y = 2x + 9 will cut y-axis when x = 0

Now if x = 0, then 3y = 2 × 0 + 9

⇒ 3y = 0 + 9 = 9 ⇒ y = \(\frac { 9 }{ 3 }\) = 3

∴ It will cut y-axis at (0, 3)