##### Parallel Lines and Perpendicular Lines

Two non-vertical lines are parallel if and only if they have the same slope. This fact is fairly intuitive in that two lines that don't both ascend or both descend at equal rates will inevitably intersect (as will two lines, one of which ascends and the other of which descends).

A less intuitive, but equally true, fact is that two lines, neither of which is vertical, are perpendicular if and only if they have opposite reciprocal slopes. For example, if a line has a slope of \(-\frac{9}{11}\text{,}\) than any line perpendicular to that line has a slope of \(\frac{11}{9}\)

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Example6.6.1

Determine the slope-intercept equation of the line that passes through the point \((9,-2)\) and is parallel to the line with equation \(x-3y=14\text{.}\)

SolutionWe begin by determining the slope of the line \(x-3y=14\text{.}\) We can do that by writing the equation in slope-intercept form, which essentially entails isolating \(y\text{.}\)

\begin{align*}
x-3y\amp=14\\
x-3y\subtractright{x}\amp=14\subtractright{x}\\
-3y\amp=-x+14\\
\multiplyleft{-\frac{1}{3}}-3y\amp=\multiplyleft{-\frac{1}{3}}(-x+14)\\
y=\frac{1}{3}x-\frac{14}{3}
\end{align*}

From the equation \(y=\frac{1}{3}x-\frac{14}{3}\) we can identify that the slope of the line \(x-3y=14 \) is \(\frac{1}{3}\text{.}\) Since the line whose equation we seek is parallel to \(x-3y=14\text{,}\) and parallel lines have equal slope, the slope of the line we seek is also \(\frac{1}{3}\text{.}\)

The line we are tasked to identify has a slope of \(\frac{1}{3}\) and passes through the point \((9,-2)\text{.}\) Using this information in the point-slope form of the linear equation and simplifying the result, the slope-intercept equation for the line is determined as follows.

\begin{gather*}
y-y_1=m(x-x_1)\\
y-(-2)=\frac{1}{3}(x-9)\\
y+2=\frac{1}{3}x-3\\
y+2\subtractright{2}=\frac{1}{3}x-3\subtractright{2}\\
y=\frac{1}{3}-5
\end{gather*}

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Example6.6.2

Let's determine the slope-intercept equation of the line that passes through the point \((-8,16)\) that is perpendicular to the line that passes through the points shown in TableĀ 6.6.3.

\(x\) |
\(y\) |

\(-4\) |
\(7\) |

\(10\) |
\(-5\) |

Table6.6.3Points on a line

SolutionWe begin by determining the slope of the line that passes through the points shown in TableĀ 6.6.3. Letting \((x_1,y_1)\) be the ordered pair \((-4,7)\) and \((x_2,y_2)\) be the ordered \((10,-5)\text{,}\) we calculate the slope as follows.

\begin{align*}
m\amp=\frac{y_2-y_1}{x_2-x_1}\\
\amp=\frac{-5-7}{10-(-4)}\\
\amp=\frac{-12}{14}\\
\amp=-\frac{6}{7}
\end{align*}

Since the line we seek is perpendicular to a line with a slope of \(-\frac{6}{7}\text{,}\) and perpendicular lines have opposite reciprocal slopes, the line we seek has a slope of \(\frac{7}{6}\text{.}\)

Using the slope of \(\frac{7}{6}\) and the ordered pair \((-8,16)\text{,}\) we determine the slope intercept equation of the line we seek as follows.

\begin{gather*}
y-y_1=m(x-x_1)\\
y-16=\frac{7}{6}(x-(-8))\\
y-16=\frac{7}{6}(x+8)\\
y-16=\frac{7}{6}x+\frac{28}{3}\\
y-16\addright{16}=\frac{7}{6}x+\frac{28}{3}\addright{\frac{48}{3}}\\
y=\frac{7}{6}x+\frac{76}{3}
\end{gather*}

##### Horizontal Lines

The equation of a horizontal line can be determined in the same manner as any other non-vertical line, although once that's done a generalization can greatly simplify the process.

Let's consider the horizontal line shown in FigureĀ 6.6.4. Let \((x_1,y_1)\) be the ordered pair \((-4,-3)\) and \((x_2,y_2)\) be the ordered pair \((2,-3)\text{.}\) We calculate the slope below.

Figure6.6.4A horizontal line

\begin{align*}
m\amp=\frac{y_2-y_1}{x_2-x_1}\\
\amp=\frac{-3-(-3)}{2-(-4)}\\
\amp=\frac{0}{6}\\
\amp=0
\end{align*}

By observation, the \(y\)-intercept of the line is \((0,-3)\text{.}\) That coupled with the slope of \(0\) gives us the following.

\begin{align*}
y\amp=mx+b\\
y\amp=0 \cdot x+(-3)\\
y\amp=-3
\end{align*}

Now that we have the equation \(y=-3\text{,}\) an appropriate response might be "d'oh!" By observation, the defining property of the points on the line is that the \(y\)-coordinates are all \(-3\)

In general, all horizontal lines have a slope of \(0\text{.}\) In addition, a horizontal line has an equation of form \(y=k\) where \(k\) is the \(y\)-coordinate common to every point on the line.

##### Vertical Lines

Consider the vertical line shown in FigureĀ 6.6.5. When we try to calculate the slope of the line, things go amok. For example:

\begin{align*}
m\amp=\frac{y_2-y_1}{x_2-x_1}\\
\amp=\frac{5-(-3)}{2-2}\\
\amp=\frac{8}{0}
\end{align*}

Figure6.6.5A vertical line

There's a phrase we attach to the last expression āuh oh! Division by \(0\) is never a good thing. Division by zero is always the result when one applies the slope formula to a vertical line, and for that reason we say that a vertical line has no slope, or that the slope is undefined.

We can see in FigureĀ 6.6.5 that every point on the line has an \(x\)-coordinate of \(2\text{,}\) so the equation of the line must be \(x=2\text{.}\) In general, vertical lines have equations of form \(x=h\text{,}\) where \(h\) is the \(x\)-coordinate of every point on the line

So, unlike determining equations of non-vertical, non-horizontal lines, determining the equation of a horizontal or vertical line is simply a matter of writing down the equation.

For example, if asked to determine the equations for horizontal and vertical lines that pass through the point \((-3,7)\text{,}\) an appropriate response would be as follows.

The horizontal line has equation \(y=7\) and the vertical line has equation \(x=-3\text{.}\)

If, in the moment, you can't remember which equation form is horizontal and which form is vertical, just sketch the line and ask yourself "which coordinate is always the same, \(x\) or \(y\text{?}\)"