Slope of a line: Problems with Solutions

Solution:
To get the slope of the straight line we need two points
$(x_{1},y_{1});(x_{2},y_{2})$ that the line passes through.

Points A and B have coordinates $(-3,-4)$ and $(0,2)$ and belong to the line.
The line slope formula is $m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

The line slope is $m=\frac{2-(-4)}{0-(-3)}=\frac{6}{3}$

$m=2$

Solution:
We need two points that a line passes through in order to find the slope of the line.
The two points are A and B with coordinates $(-2,5)$ and $(2,0)$, respectively.
The line slope formula is $m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

We calculate $m=\frac{0-(5)}{2-(-2)}=\frac{-5}{4}=-\frac{5}{4}$

Solution:
We need two points that a line passes through in order to find the slope of the line.
The two points are A and B with coordinates $(-4,4), (4,4)$ respectively.
The line slope formula is $m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

The line slope is $m=\frac{4-(4)}{4-(-4)}=\frac{0}{8}=0$

Solution:
The line slope is given by the formula $m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

$m=\frac{-1-3}{-1-1}=\frac{-4}{-2}=2$

Solution:
We will find the slope of the line using the formula
$m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

$m=\frac{3-(0)}{2-2}=\frac{3}{0}$

The graph shows us that it is a vertical line, in this type of line its $x$ coordinate is a constant.
The equation of a vertical line is a special equation determined as $x=x_{0}$ where $x_0$ is the constant value of $x$ for all points of this line. In this special type of line we say that its slope is infinite, due to the relation between the slope and its angle of inclination with respect to the positive direction
of the $x$-axis, $m=tan(90^{\circ})=\infty$
Thus, the slope of this vertical line is infinite.

Solution:
We must solve the equation for y.
$y=5x+2+1\Longrightarrow y=5x+3$
If we look at the point-slope equation of a line $y=mx+b$, we can conclude that the slope of this straight line is
$m=5$

Solution:
We must solve the equation for y.
$2y=3x+5\Longrightarrow y=\frac{3}{2}x+\frac{5}{2}$.
Let we look at the point-slope equation of a line $y=mx+b$.
We can conclude in this case that the slope of this straight line is
$m=\frac{3}{2}$

Solution:
In the equation $x+3y+5=0$ we clear the variable $y \Longrightarrow 3y=-x-5\Longrightarrow y=-\frac{1}{3}x-\frac{5}{3}$
So we observe that $m_{1}=3,m_{2}=-\frac{1}{3}$
Now we calculate the multiplication $m_{1} \cdot m_{2}=3\cdot(-\frac{1}{3})$

$m_{1} \cdot m_{2}=-1$

We perceive that the two lines are perpendicular.
So:

"Two straight lines are perpendicular if and only if the multiplication of their slopes is equal to $-1$"

Solution:
We must solve the equation $-6x+2y+7=0$ for $y$.
$2y=6x-7\Longrightarrow y=3x-\frac{7}{2}$
So we observe that $m_{1}=3,m_{2}=3$.

Let's graph the lines.
They are parallel.
It is possible to demonstrate the following relationship between the parallel lines.
Two straight lines are parallel if and only if their slopes are equal.

Solution:
We can use the point-slope form of the equation: $y-y_{1}=m(x-x_{1})$ where

$m=3$ is the slope and, $(x_{1},y_{1}) = (2, 5)$ is the point that lie to the line.

$y-5=3(x-2) \\ \Longrightarrow y=3x-6+5 \\ \Longrightarrow \fbox{y=3x-1}$

Solution:
We must find the slope of the straight line using the slope equation $m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

$m=\frac{-2-3}{1-(-2)}=\frac{-5}{3}\Longrightarrow m=-\frac{5}{3}$
Now we will use the point-slope equation to find the equation of the line
$y-y_{0}=m(x-x_{0})$ where $(x_{0},y_{0})$ is any point that belongs to the line
$y-3=-\frac{5}{3}(x-(-2)) \Longrightarrow y-3=-\frac{5}{3}x-\frac{10}{3}\Longrightarrow $
$y=-\frac{5}{3}x-\frac{10}{3}+3=-\frac{5}{3}x-\frac{1}{3}$
Thus $y=-\frac{5}{3}x-\frac{1}{3}$