Equations of Parallel and Perpendicular Lines | Geometry | LevelKite

You already know how to compare slopes and how to write an equation in point-slope form. Now you will put those skills together. You start with a line and a point, and you write the equation of a new line through that point. The new line must be parallel to the given line, meaning the two lines lie in the same plane and never meet, or perpendicular to it, meaning the two lines meet at a right angle.

Two things decide the new line: its direction and one point on it. Slope is the change in $y$ over the change in $x$ , for a nonzero change in $x$ , and it describes the direction of any line that is not vertical. A horizontal line runs straight left and right and has slope $0$ . A vertical line runs straight up and down and has undefined slope, because its change in $x$ is $0$ .

Many lines can share one direction, so the point is what picks out a single line. The given line supplies the direction, and the given point supplies the place.

Key concept

Find the slope, then use the point

From the given line, find the slope the new line must have. If the new line is not vertical, put that slope $m$ and the point $(x_1, y_1)$ into point-slope form.

Parallel Slope Criterion: two different nonvertical lines are parallel exactly when their slopes are equal.
Perpendicular Slope Criterion: two nonvertical lines are perpendicular exactly when the product of their slopes is $-1$ .

A vertical line and a horizontal line are perpendicular without a product, because a vertical line has no slope to multiply.

y - y_1 = m(x - x_1)

The words exactly when pack two statements into one, and you use both. Take parallel lines:

Forward direction: if two different nonvertical lines are parallel, then their slopes are equal. This tells you what slope the new line must have.
Converse: if two different nonvertical lines have equal slopes, then they are parallel. This lets you confirm that the line you wrote really is parallel.

The word different matters. If the point were on the given line, using the same slope would only give you the given line again. The perpendicular criterion splits into two directions the same way, with a slope product of $-1$ in place of equal slopes.

Worked example

Example 1: A parallel line

Given the line $y = 2x + 1$ and the point $(3, 1)$ , write an equation of the line through the point that is parallel to the given line.

Read the given slope. The equation is in slope-intercept form, y = mx + b, where m is the slope. So the slope is 2, and because the line has a slope, it is not vertical.
$y = 2x + 1 \quad\Rightarrow\quad m_{\text{given}} = 2$
The new line cannot be vertical, because a vertical line would cross the given line. So both lines are nonvertical, and the forward direction of the Parallel Slope Criterion applies: if two different nonvertical lines are parallel, then their slopes are equal. The new slope is 2.
$m_{\text{new}} = m_{\text{given}} = 2$
Substitute m = 2, x1 = 3 and y1 = 1 into point-slope form (Substitution Property of Equality).
$y - 1 = 2(x - 3)$
Simplify. Use the Distributive Property on the right side, then add 1 to both sides (Addition Property of Equality).
$y - 1 = 2x - 6 \quad\Rightarrow\quad y = 2x - 5$
Check. Put x = 3 into each equation. The new line gives y = 1, so it passes through the point. The given line gives y = 7, so the point is not on the given line and the two lines are different. Both lines have slope 2, so the converse of the Parallel Slope Criterion tells you they are parallel.
$2(3) - 5 = 1, \qquad 2(3) + 1 = 7$

Answer

The line through $(3, 1)$ parallel to $y = 2x + 1$ has the equation $y - 1 = 2(x - 3)$ , which simplifies to $y = 2x - 5$ .

Line A is the given line

y = 2x + 1

. Line B is the new line

y = 2x - 5

through the point

(3, 1)

.

In the figure, the horizontal axis is the $x$ -axis and the vertical axis is the $y$ -axis. Line A is the given line, and Line B is the new line through $(3, 1)$ . Each equation says $y$ rises $2$ units for every $1$ unit $x$ increases, which is what slope $2$ means.

The picture suggests that the lines never meet, but a picture is not a reason. The equations give the reason. For any $x$ , subtract Line B's $y$ -value from Line A's: $(2x + 1) - (2x - 5) = 6$ . The difference is always $6$ and never $0$ , so no point lies on both lines. That is exactly the definition of parallel lines.

Worked example

Example 2: A perpendicular line

Given the line $2x + 3y = 6$ and the point $(4, 1)$ , write an equation of the line through the point that is perpendicular to the given line.

Solve the given equation for y so you can read its slope. Subtract 2x from both sides (Subtraction Property of Equality), then divide both sides by the nonzero number 3 (Division Property of Equality). The result has the form y = mx + b, so the slope is -2/3. The given line is neither vertical nor horizontal.
$3y = -2x + 6 \quad\Rightarrow\quad y = -\tfrac{2}{3}x + 2$
Let n be the slope of the new line. The new line cannot be vertical, because a vertical line is perpendicular only to horizontal lines, and the given line is not horizontal. So both lines are nonvertical, and the forward direction of the Perpendicular Slope Criterion applies: if two nonvertical lines are perpendicular, then the product of their slopes is -1.
$-\tfrac{2}{3} \cdot n = -1$
Multiply both sides by -3/2 (Multiplication Property of Equality). This number is the reciprocal of -2/3, meaning the number whose product with -2/3 is 1.
$\left(-\tfrac{3}{2}\right)\left(-\tfrac{2}{3}\right) n = (-1)\left(-\tfrac{3}{2}\right) \quad\Rightarrow\quad n = \tfrac{3}{2}$
Substitute the slope 3/2 and the point (4, 1) into point-slope form (Substitution Property of Equality).
$y - 1 = \tfrac{3}{2}(x - 4)$
Use the Distributive Property on the right side, then add 1 to both sides (Addition Property of Equality).
$y - 1 = \tfrac{3}{2}x - 6 \quad\Rightarrow\quad y = \tfrac{3}{2}x - 5$
Check. Put x = 4 into the new equation to get y = 1, so the new line passes through the point. The slopes multiply to -1, so the converse of the Perpendicular Slope Criterion applies: if the product of the slopes of two nonvertical lines is -1, then the lines are perpendicular.
$\tfrac{3}{2}(4) - 5 = 1, \qquad -\tfrac{2}{3} \cdot \tfrac{3}{2} = -1$

Answer

The line through $(4, 1)$ perpendicular to $2x + 3y = 6$ has the equation $y - 1 = \frac{3}{2}(x - 4)$ , which simplifies to $y = \frac{3}{2}x - 5$ .

Line A is the given line

2x + 3y = 6

. Line B is the new line

y = \frac{3}{2}x - 5

through the point

(4, 1)

.

As before, the horizontal axis is the $x$ -axis and the vertical axis is the $y$ -axis. Line A is the given line and Line B is the new line. From $(0, 2)$ to $(3, 0)$ on Line A, the change in $x$ is $3$ and the change in $y$ is $-2$ , so the slope is $-\frac{2}{3}$ . From $(4, 1)$ to $(6, 4)$ on Line B, the change in $x$ is $2$ and the change in $y$ is $3$ , so the slope is $\frac{3}{2}$ . The two changes swap places and one sign changes, which is what a quarter turn does to a step on the plane. Your reason for calling the lines perpendicular is still the slope product, not how the picture looks.

The slope $\frac{3}{2}$ is the negative reciprocal of $-\frac{2}{3}$ . The negative reciprocal of a nonzero number $m$ is $-\frac{1}{m}$ , and $m \cdot \left(-\frac{1}{m}\right) = -1$ . So when the given slope is a nonzero number $m$ , the perpendicular slope is $-\frac{1}{m}$ .

Horizontal and vertical lines need extra care. For a line perpendicular to one of these, the product test cannot give you a slope: no number times $0$ equals $-1$ , and a vertical line has no slope to multiply. Instead, use this fact: a vertical line and a horizontal line are perpendicular without a product.

Point-slope form needs a slope, so it cannot write a vertical line. Use the meaning of the line instead. Every point on a vertical line has the same $x$ -coordinate, so the vertical line through $(p, q)$ is $x = p$ . Every point on a horizontal line has the same $y$ -coordinate, so the horizontal line through $(p, q)$ is $y = q$ .

For parallel lines, two different horizontal lines both have slope $0$ , so the Parallel Slope Criterion applies. Two different vertical lines have no slopes to compare, so it does not apply. They are still parallel by definition: a point cannot have two different $x$ -coordinates, so the lines never meet.

Worked example

Example 3: A horizontal given line

Given the line $y = -2$ and the point $(3, 4)$ , write an equation of the line through the point that is perpendicular to the given line.

Every point on y = -2 has y-coordinate -2, so the given line is horizontal and its slope is 0.
$m_{\text{given}} = 0$
Suppose the new line were nonvertical with slope n. The forward direction of the Perpendicular Slope Criterion would require a product of -1. But any number times 0 is 0, so no such n exists. The new line must be vertical.
$0 \cdot n = 0 \neq -1$
A vertical line has undefined slope, so point-slope form does not apply. Every point on a vertical line has the same x-coordinate, and this line passes through (3, 4), so that x-coordinate is 3.
$x = 3$
Check. The point (3, 4) has x-coordinate 3, so it lies on the new line. The new line is vertical and the given line is horizontal, and a vertical line and a horizontal line are perpendicular without a product.

Answer

The line through $(3, 4)$ perpendicular to $y = -2$ is the vertical line with the equation $x = 3$ .

The new line through the point (p, q), where the point is not on the given line

Given line	Parallel line through the point	Perpendicular line through the point
Not vertical, slope $m \neq 0$	Slope $m$ : $y - q = m(x - p)$	Slope $-\frac{1}{m}$ : $y - q = -\frac{1}{m}(x - p)$
Horizontal, slope $0$	Horizontal: $y = q$	Vertical: $x = p$
Vertical, undefined slope	Vertical: $x = p$	Horizontal: $y = q$

Takeaway: The given line tells you the direction of the new line, and the point tells you where it sits. For two different nonvertical lines, equal slopes mean parallel. For two nonvertical lines, a slope product of $-1$ means perpendicular. A vertical line and a horizontal line are perpendicular without a product, so those cases come from the meaning of the lines rather than from multiplying slopes.