SChand CLASS 9 Chapter 20 Coordinates and Graphs of Simultaneous Linear Equation Exercise 20(B)

 EXERCISE 20 B

Question 1

Ans: (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) linear

(iv) y -

Question 2

Ans: AS $x=3$ is a line parallel is $y-a x$ is at a distance af 3 unit Now the equation $x=3$ can be written as $x+0 y=3$ Giving ary values of $y$ as $1,2,3$, etc we get $x=3$

 Now plot the points

$(3,1)(3,2) \cdot(3,31)$ ----- on the graph and 

Join them to get the', required line.

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(ii) As y = -4 is a line parallel to x- axis at a distance of -4 units . we can write this equation as OX + Y = 4 

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

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

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(v) $\quad y=-3 x$

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

$\begin{array}{|c|c|c|c|}\hline x & 1 & 0 & -1 \\\hline y & -3 & 0 & 9 \\\hline\end{array}$

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

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(vi)$y=x+1$

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

$\begin{array}{|c|c|c|c|}\hline x & 1 & 0 & -1 \\y & 2 & 1 & 0 \\\hline\end{array}$

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

(vii) 

$\begin{aligned} & 2 x+y=14 \\ \Rightarrow & y=14-2 x \end{aligned}$

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

$\begin{array}{|c|c|c|c|}\hline x & 5 & 6 & 7 \\\hline y & 4 & 2 & 0 \\\hline\end{array}$

Now plot the points $(5,4)(6,2)$ and $(7,0) \mathrm{cm}$ the graph and Join them to get the required line.

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(ix)$x=3 y+1$

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

$\begin{array}{|c|c|c|c|}\hline x & 1 & 4 & -2 \\\hline y & 0 & 1 & -1 \\\hline\end{array}$

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

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Question 3

Ans: $y=3 x-7$

giving some suitable value to $x$, we gut the corresponding value of $y$ as shown below:

$\begin{array}{|c|c|c|c|}\hline x & 0 & 1 & 2 \\\hline y & -4 & -1 & 2 \\\hline\end{array}$

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

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(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

So if x =-1 then y = -7

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

meeting at 3

So $x=3, y=5$

Question 4

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

(i) So $y=5 x+1$ cut the $y-9 \times$ is, when $x=0$, when $x=0$, then $y=5 \times 0+1=0+1=1$

So $y=1$

So it will cut $y$-axis at $(0,1)$

(ii) $y=3 x-7$ will cut the $y$-axis when $x=0$ Now if $x=0$, then $y=3 \times 0-7=0-7$.

So it will cut $y$-axis at $(0,-7)$

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

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

So it will cut $y$-axis at $(0,5)$

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

Now if $x=0$ then $3 y=2 \times 0+9$

$\Rightarrow 3 y=0+9=9 \Rightarrow y=\frac{9}{5}=3$

So it will cut $y$-axis at $(0 ; 3)$