Graphs of linear equations - class-XI

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

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

In general, $x=a$ is a line lies on $x$ axis and parallel to $y$ axis and $y=b$ is a line lies on $y$ axis and parallel to $x$ axis.

The line will have a constant value of $x=7$,i.e, only $y$ will vary so it is parallel to $y-axis$.

Using the section formula, find the coordinates of the point dividing the segment joining (2,3) and (-1,2) in the ratio 3:4. Substituting these coordinates into the equation x + 2y = k yields k = 41/7.

We can write the equation in another form $4(x-3)+3(y+4)=120$
Now, we make a table

 We observe a patter that $x$ reduces by $3$ and $y$ increases by $4$. But both $x$ and $y$ cannot be negative or $0$. So, the value of $x=0$ and $y=0$ is ruled out. Hence, nine such positive integer solutions are possible.
Alternately, we can write the given equation as $4(x+3)+3(y-4)=120$. We get the same values of $x$ and $y$

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 $(0,y)$. 

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.

Statement I: Substituting (8, -1) into x + 2y = 6 gives 8 + 2(-1) = 6, which is true. Statement II is a standard definition of a solution to an equation.

Substitute the point (4, 2), where x = 4 and y = 2, into the equation 5x + a y = 28. This gives 5(4) + a(2) = 28, which simplifies to 20 + 2a = 28. Subtracting 20 gives 2a = 8, so a = 4.