Slope of a line: Problems with Solutions

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

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

The line slope is [tex]m=\frac{2-(-4)}{0-(-3)}=\frac{6}{3}[/tex]

[tex]m=2[/tex]

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 [tex](-2,5)[/tex] and [tex](2,0)[/tex], respectively.
The line slope formula is [tex]m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}[/tex]

We calculate [tex]m=\frac{0-(5)}{2-(-2)}=\frac{-5}{4}=-\frac{5}{4}[/tex]

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 [tex](-4,4), (4,4)[/tex] respectively.
The line slope formula is [tex]m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}[/tex]

The line slope is [tex]m=\frac{4-(4)}{4-(-4)}=\frac{0}{8}=0[/tex]

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

[tex]m=\frac{-1-3}{-1-1}=\frac{-4}{-2}=2[/tex]

Solution:
We use the slope formula [tex]m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}[/tex]

[tex]m=\frac{3-0}{2-2}=\frac{3}{0}[/tex]

Division by zero is not defined, so the slope cannot be calculated.
Both points have the same [tex]x[/tex]-coordinate, so the line is vertical and its equation is [tex]x=2[/tex].
A vertical line makes a [tex]90^{\circ}[/tex] angle with the [tex]x[/tex]-axis, and [tex]\tan 90^{\circ}[/tex] is not defined.

Thus, the slope of a vertical line is undefined.

Solution:
[tex]4y = -2x + 6[/tex]

[tex]y = -\frac{2}{4}x + \frac{6}{4}[/tex]

[tex]y = -\frac{1}{2}x + \frac{3}{2}[/tex]

The slope of the given line is [tex]-\frac12[/tex].

Solution:
We solve the equation for [tex]y[/tex].
[tex]y=5x+2+1\Longrightarrow y=5x+3[/tex]
Comparing with the slope-intercept form [tex]y=mx+b[/tex], we conclude that the slope of this line is
[tex]m=5[/tex]

Solution:
We solve the equation for [tex]y[/tex].
[tex]2y=3x+5\Longrightarrow y=\frac{3}{2}x+\frac{5}{2}[/tex]
This is the slope-intercept form [tex]y=mx+b[/tex], so the slope of this line is
[tex]m=\frac{3}{2}[/tex]

Solution:
In the equation [tex]x+3y+5=0[/tex] we solve for [tex]y[/tex]:
[tex]3y=-x-5\Longrightarrow y=-\frac{1}{3}x-\frac{5}{3}[/tex]
So [tex]m_{1}=3[/tex] and [tex]m_{2}=-\frac{1}{3}[/tex].
Now we calculate the product [tex]m_{1} \times m_{2}=3\times\left(-\frac{1}{3}\right)[/tex]

[tex]m_{1} \times m_{2}=-1[/tex]

We notice that the two lines are perpendicular.
So:

"Two non-vertical lines are perpendicular if and only if the product of their slopes is equal to [tex]-1[/tex]."

Solution:
We solve the equation [tex]-6x+2y+7=0[/tex] for [tex]y[/tex]:
[tex]2y=6x-7\Longrightarrow y=3x-\frac{7}{2}[/tex]
So [tex]m_{1}=3[/tex] and [tex]m_{2}=3[/tex].

The graph shows that the lines are parallel.
This illustrates the following property of parallel lines:
Two different non-vertical lines are parallel if and only if their slopes are equal.

Solution:
We can use the point-slope form of the equation of a line: [tex]y-y_{1}=m(x-x_{1})[/tex],
where [tex]m=3[/tex] is the slope and [tex](x_{1},y_{1})=(2,5)[/tex] is a point on the line.

[tex]y-5=3(x-2)[/tex]
[tex]\Longrightarrow y=3x-6+5[/tex]
[tex]\Longrightarrow \boxed{y=3x-1}[/tex]

Solution:
We find the slope of the line using the slope formula [tex]m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}[/tex]

[tex]m=\frac{-2-3}{1-(-2)}=\frac{-5}{3}\Longrightarrow m=-\frac{5}{3}[/tex]

Now we use the point-slope form to find the equation of the line:
[tex]y-y_{0}=m(x-x_{0})[/tex], where [tex](x_{0},y_{0})[/tex] is any point on the line. We take [tex](-2,3)[/tex].

[tex]y-3=-\frac{5}{3}(x-(-2))\Longrightarrow y-3=-\frac{5}{3}x-\frac{10}{3}[/tex]
[tex]y=-\frac{5}{3}x-\frac{10}{3}+3=-\frac{5}{3}x-\frac{1}{3}[/tex]

Thus [tex]y=-\frac{5}{3}x-\frac{1}{3}[/tex]