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

   EXERCISE 20 C

Question 1

Ans:

$\begin{aligned} & 3 y-2 x=7 \\ \Rightarrow & 3 y=7+2 x \\ \Rightarrow & y=\frac{7+2 x}{3} \end{aligned}$

Giving some different values to x, we get corresponding value of if as given below 

$\begin{array}{|c|c|c|c|}\hline x & 1 & 4 & -2 \\\hline y & 3 & 5 & 1 \\\hline\end{array}$

Now plot the points $(1,3),(4,5)$ and $(-2,1)$ on the graph and Join them to get a line similarly in equation.

$\begin{aligned} 5 x &+3 y=-7 \\ \Rightarrow 5 x &=-7-3 y \\ x &=-\frac{7-3 y}{5} \\ &=-\left(\frac{7+3 y}{5}\right) \end{aligned}$

$\begin{array}{|c|c|c|c|}\hline x & -2 & -5 & 1 \\\hline y & 1 & 6 & -4 \\\hline\end{array}$

Now plot the points (-2,1)(-5,6) and (1,-4) on the graph and join them to get another line. 

we see than two lines intersect 

So x = -2 , y= 1

Question 2

Ans:$2 x+3 y=13 \Rightarrow 2 x=13-3 y$

$\Rightarrow x=\frac{13-3 y}{2}$

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

$\begin{array}{|c|c|c|c|}\hline x & 5 & 2 & -1 \\\hline y & 1 & 3 & 5 \\\hline\end{array}$

Now plot the points $(5,1),(2,3)$ and $(-1,5)$ on the graph and join them to get a line similarly in equation 

$5 x-2 y=4$

$\Rightarrow 5 x=4+2 y$

$\Rightarrow x=\frac{4+2 y}{5}$

$\begin{array}{|l|l|l|l|}\hline x & 2 & 0 & -2 \\\hline y & 3 & -2 & -7 \\\hline\end{array}$

Now plot the points $(2,3),(0,-2)$ and $(-2,-7)$ on the graph and join them to get another line we see that the two lines intersect each other at the points (2,3)

so x =2 , y=3

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Question 3

Ans: $5 x+y=-3 \Rightarrow y=-3-5 x$ $\Rightarrow y=-(3+5 x)$

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

$\begin{array}{|c|c|c|c|}\hline x & 0 & -1 & -2 \\\hline y & -3 & 2 & 7 \\\hline\end{array}$

Now plot the points $(0,-3),(-1,2)$ and $(-2,7)$ on the graph and join them to get a line similarly in equation $2 x=4 y-8$ 

$\Rightarrow x=\frac{3 y-8}{2}$

$\begin{array}{rrrr}x & -4 & -1 & 2 \\ y & 0 & 2 & 4\end{array}$

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

we see that these two lines intersect each other at (-1,2)

so x = -1 , y=2

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Question 6

Ans: The line passes through the points 

Now plot the points $(4,0)$ and $(0,3)$ on the graph and join them to get a line 

If the line passes through (k,1.5)

So From 1.5 on y- axis draw a perpendicular on y - axis which intersects the line joining the point $(4,0)$ and 0,3$)$ at $P$

From P , draw a perpendicular on x - axis in which x = 2

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Question 7

Ans:  If the equation y = 3x - 3

If x = 0 , then y = 3 x 0-3

0-3=-3

If x = 1 then y = $3 \times 1-3=3-3=0$

Now plot the points $(0,-3)$ and $(110) \mathrm{cm}$ the graph and Join them we act a line 

similarly in equation.

$3 x+2 y=12$

if $x=0$ then $0+2 y=12$

$\Rightarrow \quad y=\frac{12}{2}=6$

and if $x=4$ then

$\begin{aligned}&3 \times 4+2 y=12 \Rightarrow 12+2 y=12 \\&\Rightarrow 2 y=12-12=0 \\&\Rightarrow y=0\end{aligned}$

Now plot these points (0,6) and (4,0) on the graph and join them also to get another line . we see that these two lines intersect each other at the points (2,3)

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We see that a triangle is formed by these two lines and $x$-axis whose vertices are $(2,31(1.0)$ and $(4.0)$

In this triangle base $=D C=3$ cuts and altitude $A L=3$ units

So area $=\frac{1}{2} \times 3 \times 3=\frac{9}{2}$ = 4.5 sq units 

$=4.5 \mathrm{~cm}^{2}$ or $4 \frac{1}{2}$ sq.cm

Question 8

Ans: $x+y+3=0$ and $3 x-2 y+4=0$

(i) In the equation 

$x+y+3=0 \Rightarrow x=-(y+3)$ 

Given three different value to y, we get the corresponding value of x as shown below: 

$\begin{array}{|c|c|c|c|}\hline x & -3 & -2 & -1 \\\hline y & 0 & -1 & -2 \\\hline\end{array}$

Now plot these point $(-3,0)(-2,-1)$ and $(-1,-2)$ on the graph and join them to get a line 

Similarly in the equation 

$\begin{aligned} & x-2 y+4=0 \\ \Rightarrow & 3 x=2 y-4 \\ \Rightarrow & x=\frac{2 y-4}{3} \end{aligned}$

$\begin{array}{|c|c|c|c|}\hline x & -2 & 0 & 2 \\\hline y & -1 & 2 & 5 \\\hline\end{array}$

Now plot these points $(-2,-1)(0,2)$ and $(2,5)$

On the graph and join them to get another line 

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(ii) we see that there two lines intersect each other at the point $p(-2,-1)$

so $c o$-ordinates of $p$ are $(-2,-1)$

(iii) Join OP and on measuring OP. 

We get op = 2.2 

Question 9

Ans: $x-2 y=1 \text { ard } x+y=4$

In the equation $x-2 y=1$

$\Rightarrow x+1+2 y$

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

$\begin{array}{|l|l|l|l|}\hline x & 1 & 3 & 5 \\\hline y & 0 & 1 & 2 \\\hline\end{array}$

Now plot the points $(1,0)(3,1)(5,2)$ on the graph and 

Join them to get a line 

Similarly in the equation 

$x+y=4 \Rightarrow x=4-y$

$\begin{array}{|c|c|c|c|}\hline x & 4 & 3 & 2 \\\hline y & 0 & 1 & 2 \\\hline\end{array}$

Now plot these points $(4,0),(3,1)$ and $(2,2)$ on the graph and join them to get another line 

We see that these two lines intersect each other at (3,1)

So solution is x = 3 , y = 1

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Question 10

Ans: The Two equation are $2 x-y-1=0$ and $2 x$ ty $=3$ Now in the equations

$\begin{aligned}&2 x-y-1=0 \\&2 x=y+1 \\&x=\frac{y+1}{2}\end{aligned}$

Now giving three different values to y. we get the corresponding value of x as given below.

$\begin{array}{|l|l|l|l|}\hline x & 1 & 2 & 3 \\\hline y & 1 & 3 & 5 \\\hline\end{array}$

Now plot there points $(1.2)(2,3)$ and $(3.5)$ on the graph and Join them to get a line

 similarly in the equation.

$\begin{array}{|c|c|c|c|}\hline x & 2 & 3 & 4 \\\hline y & 5 & 3 & 1 \\\hline\end{array}$

Now plot Here points $(2,5)(3,3)$ and $(4,1)$ on the graph and join them to get another line

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