Rewrite the line you want to be parallel to into the $y=mx+b$ form, if needed.

$\begin{array}{r}x–y=5\,\,\,\,\,\,\,\,\,\,\,\,\,\\−y=−x+5\\y=x–5\,\,\,\,\,\,\,\end{array}$

Identify the slope of the given line.

In the equation above, $m=1$ and $b=−5$.

Since $m=1$, the slope is $1$.

To find the slope of a parallel line, use the same slope.

The slope of the parallel line is $1$.

Use the method for writing an equation from the slope and a point on the line. Substitute 1 for m, and the point $(−2,1)$ for $x$ and $y$.

$\begin{array}{l}y=mx+b\\1=1(−2)+b\end{array}$

Solve for b.

$\begin{array}{l}1=−2+b\\3=b\end{array}$

Write the equation using the new slope for m and the b you just found.

Answer

$y=x+3$

Identify the slope of the line you want to be perpendicular to.

The given line is written in $y=mx+b$ form, with $m=2$ and $b=-6$. The slope is $2$.

To find the slope of a perpendicular line, find the reciprocal, $\displaystyle \frac{1}{2}$, then the opposite, $\displaystyle -\frac{1}{2}$.

The slope of the perpendicular line is $\displaystyle -\frac{1}{2}$.

Use the method for writing an equation from the slope and a point on the line. Substitute $\displaystyle -\frac{1}{2}$ for m, and the point $(1,5)$ for $x$ and $y$.

$\displaystyle \begin{array}{l}y=mx+b\\5=-\frac{1}{2}(1)+b\end{array}$

Solve for b.

$\displaystyle \begin{array}{l}\,\,\,5=-\frac{1}{2}+b\\\frac{11}{2}=b\end{array}$

Write the equation using the new slope for m and the b you just found.

Answer

$y=-\frac{1}{2}x+\frac{11}{2}$

Rewrite the line into $y=mx+b$ form, if needed.

You may notice without doing this that $y=4$ is a horizontal line 4 units above the x-axis. Because it is horizontal, you know its slope is zero.

$\begin{array}{l}y=4\\y=0x+4\end{array}$

Identify the slope of the given line.

In the equation above, $m=0$ and $b=4$.

Since $m=0$, the slope is $0$. This is a horizontal line.

To find the slope of a parallel line, use the same slope.

The slope of the parallel line is also $0$.

Since the parallel line will be a horizontal line, its form is

$y=\text{a constant}$

Since we want this new line to pass through the point $(0,10)$, we will need to write the equation of the new line as:

$y=10$

This line is parallel to $y=4$ and passes through $(0,10)$.

Answer

$y=10$

In the equation above, $m=0$ and $b=-3$.

A perpendicular line will have a slope that is the negative reciprocal of the slope of $y=-3$, but what does that mean in this case?

The reciprocal of $0$ is $\frac{1}{0}$, but we know that dividing by $0$ is undefined.

This means that we are looking for a line whose slope is undefined, and we also know that vertical lines have slopes that are undefined. This makes sense since we started with a horizontal line.

The form of a vertical line is $x=\text{a constant}$, where every x-value on the line is equal to some constant.  Since we are looking for a line that goes through the point $(-2,5)$, all of the $x$-values on this line must be $-2$.

The equation of a line passing through $(-2,5)$ that is perpendicular to the horizontal line $y=-3$ is therefore,

$x=-2$

Answer

$x=-2$