The question asks us to consider the following equation of a line:
[tex]y=2x-4[/tex]
We are asked to find the line parallel to this line.
First of all, we need to know the condition for two lines to be parallel.
If two lines are given by the equations:
[tex]\begin{gathered} y_{}=m_1x+c_1\text{ (Line 1)} \\ y=m_2x+c_2\text{ (Line 2)} \\ \text{where,} \\ m_1,m_2=\text{slopes of lines 1 and 2 respectively} \\ c_1,c_2=y-\text{intercept of lines 1 and 2 respectively} \end{gathered}[/tex]
For the two lines to be parallel, we would have:
[tex]\begin{gathered} m_1=m_2 \\ i\mathrm{}e\text{. the slopes of both lines must be equal} \end{gathered}[/tex]
Now, that we have this, we can find the first parallel line.
This is done below:
[tex]\begin{gathered} \text{ Comparing the equation in the question to the general} \\ \text{form of a line to get the slope:} \\ y=mx+c\equiv y=2x-4 \\ \therefore m=2,c=-4 \\ \text{ Hence, the slope is m = 2} \\ \\ By\text{ the condition for parallel lines, we can s}ee\text{ that the parallel } \\ \text{ line to the equation given in the question will also have:} \\ m=2\text{ as its slope} \end{gathered}[/tex]
Therefore, the parallel line will have the form:
[tex]\begin{gathered} y=2x+c \\ \end{gathered}[/tex]
We don't know c yet. We are told that this line passes through the point (-3, -4). Hence, to get
c, we simply substitute these points into the incomplete equation above.
This is done below:
[tex]\begin{gathered} y=2x+c \\ \text{ Using the point (-3, -4),} \\ x=-3,y=-4 \\ -4=2(-3)+c \\ -4=-6+c \\ \text{add 6 to both sides} \\ -4+6=-6+6+c \\ 2=c,c=2 \end{gathered}[/tex]
Therefore, the parallel line equation is given by:
[tex]y=2x+2[/tex]
For the perpendicular lines, the condition to be used is:
[tex]\begin{gathered} m_1=-\frac{1}{m_2} \\ \text{where,} \\ m_1,m_2\text{ are the slopes of the perpendicular lines} \end{gathered}[/tex]
Since we know the slope of the line given in the question is m = 2, therefore, the slope for the line perpendicular to it is given by:
[tex]m_2=-\frac{1}{2}[/tex]
We can now compute the perpendicular line as:
[tex]\begin{gathered} y=mx+c \\ y=-\frac{1}{2}x+c \end{gathered}[/tex]
We still need to find the y-intercept (c) of the perpendicular line. In order to do this, we simply substitute the point (-3, -4) which the perpendicular line passes through into the above incomplete equation to find c.
This is done below:
[tex]\begin{gathered} y=-\frac{1}{2}x+c \\ \text{choosing point (-3, -4)} \\ x=-3,y=-4 \\ -4=-\frac{1}{2}(-3)+c \\ -4=\frac{3}{2}+c \\ \text{subtract 3/2 from both sides} \\ -4-\frac{3}{2}=\frac{3}{2}-\frac{3}{2}+c \\ \therefore c=-4-\frac{3}{2} \\ \\ c=-\frac{11}{2} \end{gathered}[/tex]
Hence, the perpendicular line equation is:
[tex]y=-\frac{1}{2}x-\frac{11}{2}[/tex]
Therefore the final answers are:
Parallel: y = 2x + 2
Perpendicular: y = -x/2 - 11/2
Plotting the graphs, they look like this:
