Parallel and Perpendicular Lines

Parallel Lines

Definition: Two lines are said to be Parallel if for every fixed point on the first line, the minimum distance from that point to the second line is the same.

A great way to visualize parallel lines is like railroad tracks. Even though railroad tracks can bend in any such manner, the distance between them is always constant. We will now prove an important result - that is two lines are parallel if and only if they have the same slope by showing that the distance between any two corresponding points is always constant.

Suppose we have two lines that have the same slopes, say $L_1: y = mx + b_1$ and $L_2: y = mx + b_2$.

Now take any x-coordinate and call it $x_1$. Plugging this into $L_1$ and $L_2$ we can get corresponding y-coordinates, that is $(x_1, mx_1 + b_1)$ lies on $L_1$ and $(x_1, mx_1 + b_2)$ lies on $L_2$. The distance between these points will simply be the difference of their y-coordinates, that is:

(1)

\begin{align} d = mx_1 + b_1 - (mx_1 + b_2) \\ d = mx_1 + b_1 - mx_1 - b_2 \\ d = b_1 - b_2 \end{align}

Now take another other x-coordinate, call it $x_2$. Plugging this into $L_1$ and $L_2$ we can get corresponding y-coordinates, that is $(x_2, mx_2 + b_1)$ lies on $L_1$ and $(x_2, mx_2 + b_2)$ lies on $L_2$. The distance between these points will be the difference of their y-coordinates once again, that is:

(2)

\begin{align} d = mx_2 + b_1 - (mx_2 + b_2) \\ d = mx_2 + b_1 - mx_2 - b_2 \\ d = b_1 - b_2 \end{align}

Perpendicular Lines

Definition: Two lines are said to be Perpendicular if their intersection creates a 90 degree angle.

For example, the following lines $y = x$ and $y = -x$ are perpendicular as 90 degree angles are formed by their intersection:

We say that two lines are perpendicular if the have negative reciprocal slopes. We will not prove this, however we will note that "negative reciprocal" means that we take a number, multiply it by negative 1 and then flip the numerator and denominator. For example, the negative reciprocal of 4 is -1/4. The negative reciprocal of -1/2 is 2.

For example, the lines $y = 2x + 1$ and $y = -\frac{1}{2}x + 4$ are perpendicular since their slopes are negative reciprocals of each other.

Since the distance is the same, the lines must be parallel.

Example Questions

• 1. If two lines are parallel and have the same y-intercept, what can be said about the lines?

• 2. Find the equation of a line that is parallel to $y = 2x + 3$ and passes through the point $(2, 2)$.

• 3. Find the equation of a line that is perpendicular to $y = 4x + 1$ and has a $y$-intercept at $(0, 4)$.

• 4. Let $L_1: ax + by + c = 0$ be the equation of a line. Give a general form equation for a line that is perpendicular to $L_1$ and has the same $y$-intercept.

Solutions

See solutions

• 1. If two lines are parallel and have the same y-intercept, then the lines are the same.

• 2. We note that this line must have a slope of 2 and pass through the point $(x, y) = (2, 2)$. We must hence solve for the y-intercept by plugging in and solving for $b$ as follows:

(3)

\begin{align} y = mx + b 2 = 2(2) + b \\ 2 = 4 + b -2 = b \end{align}

• Therefore, we know $b = -2$ and $m = 2$, so $y = 2x - 2$ is a line that has a slope of 2 and passes through the point $(2, 2)$.

• 3. We note that this line must have a negative reciprocal slope, so $m = -\frac{1}{4}$. Our y-intercept is $(0, 4)$, so $b = 4$. Therefore, plugging these values back into $y = mx + b$ we get that $y = -\frac{1}{4}x + 4$ is a line that is perpendicular to our original and has a y-intercept of $(0, 4)$.

• 4. We first take $ax + by + c = 0$ and rewrite it in slope-intercept form, that is:

(4)

\begin{align} y = -\frac{a}{b}x - \frac{c}{b} \end{align}

• Therefore, our line must have $m = \frac{b}{a}$ to be perpendicular, and $b = -\frac{c}{b}$ to have the same y-intercept, and thus our line is:

(5)

\begin{align} y = \frac{b}{a}x - \frac{c}{b} \end{align}

• Now all we have to do is convert this to general form as follows:

(6)

\begin{align} y = \frac{b}{a}x - \frac{c}{b} \\ 0 = \frac{b}{a}x -y - \frac{c}{b} \\ 0 \cdot a = \left ( \frac{b}{a}x -y - \frac{c}{b} \right ) \cdot a \\ 0 = bx - ay - \frac{ac}{b} \\ 0 \cdot b = \left ( bx - ay - \frac{ac}{b} \right ) \cdot b \\ 0 = b^2x - aby - ac \\ \end{align}