Rewrite the line you want to be parallel to into the [latex]y=mx+b[/latex] form, if needed.

[latex]\begin{array}{r}x–y=5\,\,\,\,\,\,\,\,\,\,\,\,\,\\−y=−x+5\\y=x–5\,\,\,\,\,\,\,\end{array}[/latex]

Identify the slope of the given line.

In the equation above, [latex]m=1[/latex] and [latex]b=−5[/latex].

Since [latex]m=1[/latex], the slope is [latex]1[/latex].

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

The slope of the parallel line is [latex]1[/latex].

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

[latex]\begin{array}{l}y=mx+b\\1=1(−2)+b\end{array}[/latex]

Solve for b.

[latex]\begin{array}{l}1=−2+b\\3=b\end{array}[/latex]

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

Answer

[latex]y=x+3[/latex]

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

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

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

The slope of the perpendicular line is [latex]\displaystyle -\frac{1}{2}[/latex].

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

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

Solve for b.

[latex]\displaystyle \begin{array}{l}\,\,\,5=-\frac{1}{2}+b\\\frac{11}{2}=b\end{array}[/latex]

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

Answer

[latex]y=-\frac{1}{2}x+\frac{11}{2}[/latex]

Rewrite the line into [latex]y=mx+b[/latex] form, if needed.

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

[latex]\begin{array}{l}y=4\\y=0x+4\end{array}[/latex]

Identify the slope of the given line.

In the equation above, [latex]m=0[/latex] and [latex]b=4[/latex].

Since [latex]m=0[/latex], the slope is [latex]0[/latex]. 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 [latex]0[/latex].

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

[latex]y=\text{a constant}[/latex]

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

[latex]y=10[/latex]

This line is parallel to [latex]y=4[/latex] and passes through [latex](0,10)[/latex].

Answer

[latex]y=10[/latex]

In the equation above, [latex]m=0[/latex] and [latex]b=-3[/latex].

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

The reciprocal of [latex]0[/latex] is [latex]\frac{1}{0}[/latex], but we know that dividing by [latex]0[/latex] 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 [latex]x=\text{a constant}[/latex], where every x-value on the line is equal to some constant.  Since we are looking for a line that goes through the point [latex](-2,5)[/latex], all of the [latex]x[/latex]-values on this line must be [latex]-2[/latex].

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

[latex]x=-2[/latex]

Answer

[latex]x=-2[/latex]