Parallel and Perpendicular Lines

Learning goals

Match slopes for parallel lines with different intercepts
Take the negative reciprocal for a perpendicular slope
Treat vertical and horizontal lines as the exception to the product rule
Classify two lines by comparing their slopes
Write the required line through a point with point-slope form

Parallel lines: equal slopes

Slope measures direction, the tilt of a line as rise over run. Two lines point the same way exactly when they climb at the same rate, so it should feel right that parallel lines have equal slopes. Suppose one line rises two units for every unit across and the other rises three. Then the steeper one is slowly pulling away from the flatter one, and sooner or later they cross. Only when the two rates match does the distance between the lines hold steady forever, and holding a steady distance is what it means to be parallel.

That picture is convincing, but the algebra makes it exact. The algebra also sorts out the one case the picture glosses over: two lines with the same slope that are secretly the same line.

Why equal slopes mean parallel#

Take two non-vertical lines and write each in slope-intercept form, $y = m_1 x + b_1$ and $y = m_2 x + b_2$ . The lines meet at any point where their $y$ -values agree, so set the right-hand sides equal and see which $x$ can satisfy both:

m_1 x + b_1 = m_2 x + b_2.

Gather the $x$ -terms on one side:

(m_1 - m_2)\,x = b_2 - b_1.

Everything now hinges on whether the slopes are equal. Suppose first that they differ, so $m_1 \neq m_2$ . Then $m_1 - m_2$ is not zero, and you may divide by it:

x = \frac{b_2 - b_1}{m_1 - m_2}.

That is one definite number, a single $x$ where the lines cross. Different slopes force exactly one crossing, so lines with different slopes are never parallel.

Now suppose the slopes are equal, $m_1 = m_2$ . Then $m_1 - m_2 = 0$ , and the equation collapses to $0 = b_2 - b_1$ . If the intercepts differ, $b_1 \neq b_2$ , this says $0$ equals a nonzero number, which is impossible: no $x$ works, the lines share no point, and they are parallel. If instead the intercepts are also equal, $b_1 = b_2$ , then every $x$ works, and the two equations describe one and the same line lying on top of itself.

So among non-vertical lines, equal slopes with different intercepts is precisely the parallel case, and it is the only way two such lines can fail to cross.

One family sits outside this argument because it has no slope to compare: the vertical lines. A vertical line has the equation $x = c$ and an undefined slope, so slope-intercept form never applies to it. Even so, any two vertical lines run straight up and down side by side and never meet, so all vertical lines are parallel to one another. You handle them by inspection rather than by matching a slope number.

Two parallel lines. Both have slope 1/2, so they climb at the same rate, and the vertical gap between them stays a constant 3 units at every x. Equal slope with different y-intercepts is the parallel case.

Worked example 1 Decide whether two lines are parallel

Are the lines $y = -3x + 5$ and $6x + 2y = 1$ parallel? Parallelism is about slope, so get both slopes into view. The first line is already in slope-intercept form, and its slope is $m_1 = -3$ .

The second line is in standard form, so solve it for $y$ to expose its slope. Subtract $6x$ from both sides, then divide every term by $2$ :

2y = -6x + 1, \qquad y = -3x + \tfrac{1}{2}.

Its slope is $m_2 = -3$ as well. The two slopes are equal, $-3 = -3$ , and the intercepts differ, $5 \neq \tfrac{1}{2}$ , so the lines are parallel and not the same line.

Check your understanding

Are the lines $y = \tfrac{2}{5}x + 1$ and $2x - 5y = 20$ parallel, perpendicular, or neither?

A Parallel B Perpendicular C Neither D They are the same line

Perpendicular lines: negative reciprocal slopes

Perpendicular is the right-angle case, and the slope rule for it is less obvious than for parallel lines. Two non-vertical lines are perpendicular exactly when their slopes are negative reciprocals: flip one slope over and change its sign to get the other. Written as an equation, the slopes satisfy

m_1 \cdot m_2 = -1, \qquad \text{equivalently} \qquad m_2 = -\frac{1}{m_1}.

So a line of slope $2$ is perpendicular to a line of slope $-\tfrac{1}{2}$ . A line of slope $-\tfrac{3}{4}$ is perpendicular to a line of slope $\tfrac{4}{3}$ , and in every case the product of the two slopes is $-1$ . This is not a rule to accept on faith. It falls out of a single idea: a perpendicular line is the original line turned a quarter turn. Turning a slope triangle a quarter turn does a predictable thing to its rise and run.

Why perpendicular slopes are negative reciprocals#

Start with a line $L_1$ that is neither horizontal nor vertical. Write its slope as a fraction, $m_1 = \dfrac{a}{b}$ , where $b$ is a horizontal run and $a$ is the matching vertical rise. Both $a$ and $b$ are nonzero, since the line tilts. Pick any point $P$ on $L_1$ . Starting at $P$ , step $b$ to the right and then $a$ upward, and you land on a second point of $L_1$ . The two steps form a slope triangle with a horizontal leg of length $b$ and a vertical leg of length $a$ .

Now rotate the whole figure a quarter turn (ninety degrees counterclockwise) about the point $P$ . Rotation is rigid and turns every direction by the same angle. So the line $L_1$ swings to a new line $L_2$ through $P$ , and that new line meets $L_1$ at a right angle. That is exactly a line perpendicular to $L_1$ , so finding the slope of $L_2$ finds the slope of the perpendicular direction.

Track the slope triangle through the turn. A quarter turn counterclockwise sends a step ” $b$ to the right” to a step ” $b$ upward”. The same turn sends a step ” $a$ upward” to a step ” $a$ to the left”. In vector form, the arrow $(b,\ a)$ that points along $L_1$ becomes the arrow $(-a,\ b)$ that points along $L_2$ . So starting again at $P$ , the rotated triangle runs $a$ units to the left and rises $b$ units up. Its run is $-a$ (leftward is negative) and its rise is $b$ , which makes the slope of $L_2$

m_2 = \frac{\text{rise}}{\text{run}} = \frac{b}{-a} = -\frac{b}{a}.

Compare that with $m_1 = \dfrac{a}{b}$ : the fraction has been flipped and its sign changed, so $m_2$ is the negative reciprocal of $m_1$ . Multiplying the two slopes confirms it directly:

m_1 \cdot m_2 = \frac{a}{b} \cdot \left(-\frac{b}{a}\right) = -\frac{ab}{ab} = -1.

The argument runs backward too. Suppose two non-vertical lines have slopes whose product is $-1$ , which forces one slope to be $-\dfrac{b}{a}$ when the other is $\dfrac{a}{b}$ . Then the same quarter turn carries the first line onto a line parallel to the second. Parallel lines lean the same way, so the two lines meet at a right angle. Perpendicular and “slopes multiply to $-1$ ” are therefore the same condition, each guaranteeing the other.

Turning a slope triangle a quarter turn. The line L1 has slope a/b, drawn here with run b = 2 and rise a = 1. Rotating its triangle ninety degrees about P swaps the legs: the run of 2 becomes a rise of 2, and the rise of 1 becomes a run of 1 to the left. The turned line L2 has slope b over negative a, that is negative b/a, and the two lines meet at a right angle.

With the rule in hand, the two jobs it does are quick: reading off a perpendicular slope, and testing whether two given lines are perpendicular.

Worked example 2 Find a perpendicular slope, then test a pair of lines

First, find the slope of any line perpendicular to $2x - 3y = 6$ . Solve for $y$ to read its slope. Subtract $2x$ and divide by $-3$ :

-3y = -2x + 6, \qquad y = \tfrac{2}{3}x - 2,

so this line has slope $\tfrac{2}{3}$ . A perpendicular line has the negative reciprocal slope: flip $\tfrac{2}{3}$ to $\tfrac{3}{2}$ and change the sign,

m_{\perp} = -\frac{3}{2}.

Second, are $y = \tfrac{2}{3}x + 1$ and $y = -\tfrac{3}{2}x + 4$ perpendicular? Multiply their slopes and check for $-1$ :

\frac{2}{3} \cdot \left(-\frac{3}{2}\right) = -\frac{6}{6} = -1.

The product is $-1$ , so yes, the two lines meet at a right angle.

Two perpendicular lines. One has slope 2 and the other has slope negative one half, and 2 times negative one half is negative 1. They cross at (2, 1), and the small square marks the right angle there. Drawing the axes to the same scale is what makes the right angle look square.

Check your understanding

What is the slope of any line perpendicular to the line through the points $(1, 2)$ and $(5, 4)$ ?

\tfrac{1}{2}

2

-2

-\tfrac{1}{2}

The horizontal and vertical case

The negative-reciprocal rule quietly assumes both lines have real-number slopes. So one pair of perpendicular lines slips through the cracks and has to be handled on its own: a horizontal line and a vertical line.

A horizontal line has slope $0$ , and a vertical line has an undefined slope. Picture the $x$ -axis and the $y$ -axis: one runs flat, the other stands straight up, and they cross at a perfect right angle. So a horizontal line is perpendicular to every vertical line, plainly and by sight. But you cannot confirm it with $m_1 \cdot m_2 = -1$ , because a vertical line has no slope number to multiply, and $0$ times “undefined” is not a meaningful product. On the horizontal side the trouble shows up differently: the negative reciprocal of $0$ would be $-\tfrac{1}{0}$ . Dividing by zero is not allowed, so slope $0$ has no negative reciprocal at all. The perpendicular partner of a horizontal line is a vertical line, and you know it because it stands upright, not because a slope product came out to $-1$ .

The takeaway is a small exception to keep in your pocket. Horizontal is perpendicular to vertical always, but that one pair is verified by the picture, not by the slope formula, which needs two ordinary slopes to work.

Testing whether two lines are parallel, perpendicular, or neither

Every classification problem reduces to the same short routine. Find the slope of each line, putting each equation into slope-intercept form or applying the slope formula to two points if that is what you are given. Then compare the two slopes: if they are equal, the lines are parallel (or the same line, when the intercepts match too). If their product is $-1$ , the lines are perpendicular. If neither the equal-slope test nor the product test holds, the lines are neither, crossing at some angle that is not a right angle.

Worked example 3 Classify three pairs of lines

Classify each pair as parallel, perpendicular, or neither.

Pair A: $y = 4x - 1$ and $y = 4x + 6$ . Both slopes are $4$ , and the intercepts differ, so the lines are parallel.

Pair B: $y = \tfrac{3}{2}x + 5$ and $2x + 3y = 9$ . Convert the second: $3y = -2x + 9$ , so $y = -\tfrac{2}{3}x + 3$ . The slopes are $\tfrac{3}{2}$ and $-\tfrac{2}{3}$ , and their product is

\frac{3}{2} \cdot \left(-\frac{2}{3}\right) = -1,

so the lines are perpendicular.

Pair C: $y = 2x + 1$ and $y = -2x + 1$ . The slopes are $2$ and $-2$ . They are not equal, so the lines are not parallel, and their product is $2 \cdot (-2) = -4$ , not $-1$ , so they are not perpendicular either. This pair is neither. Watch the trap: the slopes look related, but a sign flip alone is not the same as a negative reciprocal, which also flips the fraction.

Check your understanding

The lines $3x + 4y = 8$ and $4x - 3y = 12$ are which of the following?

A Parallel B The same line C Perpendicular D Neither

The figure below opens on Pair C, the trap from the worked example: $y = 2x + 1$ and $y = -2x + 1$ . Read the sentence under it. It reports that the slopes are different and the lines cross once, and it says nothing about a right angle, because there is not one there.

Now fix the trap by hand. Leave the first line alone and change the second to a slope of $-\tfrac{1}{2}$ : take its rise to $-1$ and its run to $2$ . A sign flip was never enough, and this is the step it was missing, the fraction turning over. The moment you arrive the sentence changes to name the right angle, and the two lines visibly square up.

Two more things are worth doing while it is in front of you. Set both lines to the same slope and watch the crossing dot disappear: parallel lines have nowhere to meet, and the figure stops claiming one. Then, holding that, walk the second line’s crossing point onto the first line’s. The picture now shows what looks like a single line, and this is the one moment where looking is not enough. The sentence is what tells you the two equations describe the same line, so every point on it satisfies both, rather than a pair that never meets.

Parallel, perpendicular, or neither?

Rise A Run A Crosses A Rise B Run B Crosses B

y = 2x + 1. y = -2x + 1. The slopes 2 and -2 are different, so they cross exactly once. They cross at (0, 1).

Two lines on one coordinate plane, one solid and one dashed, each with its own rise, run and crossing point. A dot marks where they meet, and it is absent whenever they do not.

Writing equations of parallel and perpendicular lines

The slope rules also let you build a line to specification. Given a line and a point not on it, write the line through that point that is parallel, or perpendicular, to the one you were handed. When the given line has a slope, the plan is two steps. Get the slope you need from it (the same slope for parallel, the negative reciprocal for perpendicular). Then feed that slope and the given point into point-slope form and simplify.

Worked example 4 A line through a point, parallel to a given line

Write the equation of the line through $(2, -1)$ that is parallel to $y = 3x + 4$ , in slope-intercept form.

Parallel means the same slope, and the given line has slope $3$ , so the new line also has slope $3$ . Now use point-slope form with the point $(2, -1)$ :

y - (-1) = 3(x - 2), \qquad y + 1 = 3x - 6.

Subtract $1$ from both sides to reach slope-intercept form:

y = 3x - 7.

As a check, the slope is $3$ as required, and at $x = 2$ the equation gives $y = 6 - 7 = -1$ , so the line passes through $(2, -1)$ .

Worked example 5 A line through a point, perpendicular to a given line

Write the equation of the line through $(4, 1)$ that is perpendicular to $y = 2x - 5$ , in slope-intercept form.

The given line has slope $2$ , so a perpendicular line has the negative reciprocal slope: flip $2$ to $\tfrac{1}{2}$ and change the sign, giving $m = -\tfrac{1}{2}$ . Use point-slope form with the point $(4, 1)$ :

y - 1 = -\tfrac{1}{2}(x - 4), \qquad y - 1 = -\tfrac{1}{2}x + 2.

Add $1$ to both sides:

y = -\tfrac{1}{2}x + 3.

The slope $-\tfrac{1}{2}$ is the negative reciprocal of $2$ , and at $x = 4$ the equation gives $y = -2 + 3 = 1$ , so the line runs through $(4, 1)$ as required.

Worked example 6 Perpendicular to a line given in standard form

Write the equation of the line through $(-3, 2)$ that is perpendicular to $3x + y = 7$ .

First get the slope of the given line by solving for $y$ : $y = -3x + 7$ , so its slope is $-3$ . The perpendicular slope is the negative reciprocal of $-3$ : flip to $-\tfrac{1}{3}$ and change the sign, giving $m = \tfrac{1}{3}$ . Now use point-slope form with $(-3, 2)$ :

y - 2 = \tfrac{1}{3}\bigl(x - (-3)\bigr) = \tfrac{1}{3}(x + 3).

Distribute and add $2$ :

y - 2 = \tfrac{1}{3}x + 1, \qquad y = \tfrac{1}{3}x + 3.

The slope $\tfrac{1}{3}$ times the original $-3$ gives $-1$ , confirming perpendicularity, and the line passes through $(-3, 2)$ since $\tfrac{1}{3}(-3) + 3 = 2$ .

Check your understanding

Which is the equation of the line through $(0, 3)$ that is perpendicular to $y = \tfrac{1}{2}x - 8$ ?

y = \tfrac{1}{2}x + 3

y = 2x + 3

y = -2x + 3

y = -2x - 3

Common mistakes

Flipping the sign but not the fraction for a perpendicular slope. The perpendicular of slope $3$ is $-\tfrac{1}{3}$ , not $-3$ . A negative reciprocal both flips the fraction and changes the sign, so $3 \cdot (-\tfrac{1}{3}) = -1$ , while $3 \cdot (-3) = -9$ .
Flipping the fraction but not the sign. The perpendicular of slope $\tfrac{2}{5}$ is $-\tfrac{5}{2}$ , not $\tfrac{5}{2}$ . Without the minus sign the product is $+1$ , not $-1$ , so the angle is not a right angle.
Reading a slope before solving for $y$ . In $6x + 2y = 1$ the slope is not $6$ . Solve for $y$ first to get $y = -3x + \tfrac{1}{2}$ ; only then is the coefficient of $x$ the slope.
Calling equal-slope lines “the same line”. Equal slopes with different intercepts are parallel, two separate lines. They coincide only when the intercepts match as well.
Forcing the product rule onto a horizontal and a vertical line. A vertical line has no slope number, so you cannot check $m_1 \cdot m_2 = -1$ for it. A horizontal line and a vertical line are perpendicular by sight; that pair is the one exception to the slope test.
Using the given point as the intercept. In the point-slope step, the point $(2, -1)$ is not the $y$ -intercept unless its $x$ -coordinate is $0$ . Substitute the point into point-slope form and simplify to find the real intercept.

Key takeaways

Two non-vertical lines are parallel exactly when their slopes are equal, $m_1 = m_2$ , with different $y$ -intercepts. Equal slopes and equal intercepts mean the same line.
Two non-vertical lines are perpendicular exactly when their slopes are negative reciprocals, $m_2 = -\dfrac{1}{m_1}$ , equivalently $m_1 \cdot m_2 = -1$ . Flip the fraction and change the sign.
All vertical lines $x = c$ are parallel to each other, and every horizontal line is perpendicular to every vertical line. That right angle is confirmed by sight, since a vertical line has no slope to put in the product rule, and $0$ has no negative reciprocal.
To classify two lines, find both slopes (convert to $y = mx + b$ or use the slope formula), then compare: equal for parallel, product $-1$ for perpendicular, otherwise neither.
To write a parallel or perpendicular line through a point, take the required slope (same slope, or the negative reciprocal). Then substitute that slope and the point into point-slope form and simplify.

Parallel and Perpendicular Lines

Learning goals

Parallel lines: equal slopes

Perpendicular lines: negative reciprocal slopes

The horizontal and vertical case

Testing whether two lines are parallel, perpendicular, or neither

Writing equations of parallel and perpendicular lines

Common mistakes

Practice

Multiple Choice Questions (MCQ)

Free Response Questions (FRQ)