Tag: graphs of linear equations

Put $\displaystyle x=6$ and $\displaystyle y=0$ in given equation.
$\displaystyle 2x-3y=2\times 6-3\times 0$
=12-0
=12
Hence, the value of x and y that satisfy the equation are (6,0).

The line $x=8$ or a point $(8,0)$ lies on the $x$-axis whwich is parallel to $y$-axis and at a distance of $8$ units.

Any point on $x$ axis is of the form $(x,0)$ and any point on $y$ axis is of the form $(0,y)$. 

Given $|x|+|y|$

1. For $x>0$ and $y>0$ , we have $x+y$
2. For $x>0$ and $y<0$ , we have $x-y$
3. For $x<0$ and $y>0$ , we have $-x+y$
4. For $x<0$ and $y<0$ , we have $-x-y$
5. observe that the four equations will form a square

Since, if a line at a distance $a$ from $x$-axis and parallel to $x$-axis it is given by $y=+a$ or $y=-a$
Here line is $y=5$
Distance of given line from $x$-axis is $5$ units.

Since, any line of type $x=a$ is parallel to $y$-axis and at a distance $a$ from $y$-axis.

Given, $4x-3=0$
$\Rightarrow 4x=3$
$\Rightarrow x=\dfrac 34$
Since, every line of type $x=a$ is parallel to $y$ axis.
So, option A is correct.

Let the equation of line be $y=mx+c  ...(1)$
Since, the line is parallel to $x$-axis, so slope of line, $m =0$
So, equation $(1)$ becomes $y=c$
Let the line is at a distance of $a$ from $x$-axis.
$a=c$
$\Rightarrow y=a$

The graph of the equation x = b is always a straight line parallel to y-axis.

The graph of the equation y = a is always a straight line parallel to x-axis.