Why is the product of perpendicular slopes exactly -1?

Rotating a line 90^ about a point turns a rise of a and a run of b into a rise of b and a run of -a (or -b and a). So the slope a/b becomes b/-a=-b/a, the opposite reciprocal. Multiplying the two gives a/b·(-b/a)=-1. The negative sign is what makes one line rise while the other falls.

What do I do when one line is vertical?

Do not try the multiplication test, because a vertical line has undefined slope and you cannot multiply by "undefined." Instead reason by form. A line perpendicular to the vertical line x=c is horizontal, so through the point (p,q) it is y=q. A line parallel to x=c through (p,q) is x=p. Similarly, perpendicular to y=c through (p,q) is x=p.

Should I answer in point-slope or slope-intercept form?

Follow whatever the problem or your teacher asks for. Point-slope y-y_1=m(x-x_1) is fastest to write because you just substitute. Slope-intercept y=mx+b is better when you need to graph or compare y-intercepts. They describe the same line, so converting is only a matter of distributing and adding.

Two lines have equal slopes but I got the same equation twice. Are they parallel?

No. If the slopes and the y-intercepts both match, the two descriptions are of one single line, sometimes called coincident lines. Parallel requires the lines to be distinct and never intersect. Whenever slopes come out equal, test one point from the second line in the first line's equation; if it satisfies the equation, the lines coincide.

GEOM-3.3

Slopes & Equations of Parallel and Perpendicular Lines

Learn how slope proves lines are parallel, perpendicular, or neither, and how to write equations through a point using point-slope form. Geometry 3.3.

What you'll do in this lesson

A voice-first session with the Crimsora tutor on Slopes & Equations of Parallel and Perpendicular Lines, then targeted practice and FRQs — with the tutor adapting to where you get stuck.

Start Slopes & Equations of Parallel and Perpendicular Lines free

What this lesson covers

You already know that parallel lines never meet and that perpendicular lines form right angles. In Unit 3 you proved those facts with angle pairs and transversals. Now you get a second, completely independent tool: coordinates. Once a line lives on the coordinate plane, its slope is a single number that encodes its direction — and comparing two slopes settles the parallel/perpendicular question instantly, without measuring a single angle.

This lesson has two halves. First, the test: given two lines (as equations, as graphs, or as pairs of points), decide whether they are parallel, perpendicular, or neither. Second, the construction: given one line and one point, write the equation of the new line through that point that is parallel or perpendicular to the original. The second half is where most of the homework lives, and point-slope form makes it a three-step routine you can do reliably every time.

Slope as a Direction Fingerprint

The slope of a line through

(x_1,y_1)

and

(x_2,y_2)

m=\frac{y_2-y_1}{x_2-x_1}

Slope measures steepness and direction: rise over run. Two lines that point the same way have the same slope. That is the whole idea behind the parallel test.

Parallel lines have equal slopes. If

m_1=m_2

and the lines are not the same line, they are parallel. The "not the same line" clause matters:

y=2x+1

and

y=2x+1

have equal slopes but are one line, not two parallel lines. Check the

y

-intercepts too. If slopes and intercepts both match, the lines coincide.

Perpendicular lines have slopes whose product is $-1$ . Equivalently, each slope is the opposite reciprocal of the other: if

m_1=\frac{3}{4}

, then

m_2=-\frac{4}{3}

. Notice both changes happen — flip the fraction and change the sign. Flipping only one is the single most common error in this lesson. A quick self-check: multiply.

\frac{3}{4}\cdot\left(-\frac{4}{3}\right)=-1

, so the lines are perpendicular.

Horizontal and vertical lines are the exception. A horizontal line

y=5

has slope

0

. A vertical line

x=5

has undefined slope, so you cannot multiply and get

-1

. Yet they clearly meet at a right angle. Handle this pair by recognizing the forms, not by arithmetic: every horizontal line is perpendicular to every vertical line, and two horizontals (or two verticals) are parallel.

Relationship	Slope condition
Parallel	$m_1=m_2$ , different intercepts
Perpendicular	$m_1\cdot m_2=-1$ , or one is horizontal and the other vertical
Same line	$m_1=m_2$ and same intercept
Neither	none of the above

Getting the Slope Out of Any Equation

Before you can compare slopes, you have to find them, and equations arrive in several disguises.

Slope-intercept form

y=mx+b

hands you the slope directly:

m

is the coefficient of

x

. In

y=-\frac{2}{5}x+7

, the slope is

-\frac{2}{5}

.

Standard form

Ax+By=C

hides it. Solve for

y

. From

3x+4y=12

, subtract:

4y=-3x+12

, then divide:

y=-\frac{3}{4}x+3

, so

m=-\frac{3}{4}

. A shortcut worth knowing is

m=-\frac{A}{B}

, but only use it after you have practiced the algebra enough to see why it works.

Point-slope form

y-y_1=m(x-x_1)

also displays the slope as the multiplier out front. In

y+2=\frac{1}{2}(x-6)

, the slope is

\frac{1}{2}

and the line passes through

(6,-2)

. Watch the signs:

y+2

means

y_1=-2

.

Two points require the slope formula. Subtract in the same order in both numerator and denominator. Doing

y_2-y_1

over

x_1-x_2

flips the sign and turns a parallel answer into a perpendicular-looking one.

A graph gives you slope by counting rise and run between two lattice points the line clearly passes through. Count up or down first (positive up), then right (positive right).

One trap: an equation like

2y=6x-8

is not in slope-intercept form even though it looks close. The slope is not

6

. Divide everything by

2

first to get

y=3x-4

, slope

3

. Always isolate

y

with a coefficient of exactly

1

before reading off

m

Writing the Equation Through a Given Point

This is the construction half of the objective, and it is the same three steps every time.

Step 1: find the slope of the given line. Step 2: convert that slope to the slope you need — keep it for parallel, take the opposite reciprocal for perpendicular. Step 3: substitute the new slope and the given point into point-slope form

y-y_1=m(x-x_1)

, then simplify to whatever form the problem asks for.

Example: write the line through

(4,-1)

parallel to

y=\frac{3}{2}x-8

. The slope is

\frac{3}{2}

; parallel means keep it. Then

y-(-1)=\frac{3}{2}(x-4)

, so

y+1=\frac{3}{2}x-6

and

y=\frac{3}{2}x-7

.

Same point, perpendicular instead: the slope becomes

-\frac{2}{3}

, giving

y+1=-\frac{2}{3}(x-4)

, so

y=-\frac{2}{3}x+\frac{8}{3}-1=-\frac{2}{3}x+\frac{5}{3}

.

Where students go wrong: substituting the point into the original equation instead of the new one, or reusing the original

y

-intercept. The new line almost never shares the old intercept — if it did, the two lines would cross at the

y

-axis, which parallel lines cannot do.

You can also use slope-intercept form: plug

m

and the point into

y=mx+b

and solve for

b

. For the perpendicular case above:

-1=-\frac{2}{3}(4)+b

, so

-1=-\frac{8}{3}+b

and

b=\frac{5}{3}

. Same answer. Point-slope is usually faster because it needs no solving; slope-intercept is handy when the final form must be

y=mx+b

anyway.

Always sanity-check by substituting the given point back into your final equation.

Using the Slope Criteria in Proofs and Shapes

The reason this standard exists is that slope lets you prove things about figures placed on a grid. Given four vertices, you can decide what kind of quadrilateral you have without a protractor.

Suppose

A(0,0)

B(4,2)

C(6,6)

D(2,4)

. Compute all four side slopes:

AB

has slope

\frac{2}{4}=\frac{1}{2}

;

DC

has slope

\frac{6-4}{6-2}=\frac{1}{2}

;

AD

has slope

\frac{4}{2}=2

;

BC

has slope

\frac{6-2}{6-4}=2

. Both pairs of opposite sides are parallel, so

ABCD

is a parallelogram. Is it a rectangle? Check adjacent sides:

\frac{1}{2}\cdot 2=1

, not

-1

, so there are no right angles and it is not a rectangle. Slopes cannot settle the rest, so measure the sides: all four come out to

\sqrt{20}

, which makes

ABCD

a rhombus.

This is exactly how coordinate geometry proofs are written: state the slopes, state the criterion you are using, state the conclusion. "Since

m_{AB}=m_{DC}=\frac{1}{2}

\overline{AB}\parallel\overline{DC}

by the slope criterion for parallel lines."

Slope also identifies altitudes and perpendicular bisectors. The altitude from a vertex to the opposite side is the line through that vertex whose slope is the opposite reciprocal of that side's slope — the exact construction from the previous section. The perpendicular bisector of a segment needs two ingredients: the midpoint

\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)

as the point, and the opposite reciprocal of the segment's slope.

A caution: slope tells you about direction only. Two segments with equal slopes are parallel, but they may have wildly different lengths, and they might even lie on the same line. When a proof needs congruence, you still need the distance formula. Slope answers "which way," distance answers "how far."

Key terms

Slope.: The ratio of vertical change to horizontal change between two points on a line, $m=\frac{y_2-y_1}{x_2-x_1}$ .
Parallel lines.: Coplanar lines that never intersect. In the coordinate plane, two distinct lines are parallel exactly when their slopes are equal (or both are vertical).
Perpendicular lines.: Lines that intersect at a right angle. Their slopes multiply to $-1$ , unless one is horizontal and the other vertical.
Opposite reciprocal.: The result of flipping a fraction and changing its sign; the opposite reciprocal of $\frac{a}{b}$ is $-\frac{b}{a}$ . Perpendicular slopes are opposite reciprocals.
Point-slope form.: The equation $y-y_1=m(x-x_1)$ for the line with slope $m$ through the point $(x_1,y_1)$ .
Slope-intercept form.: The equation $y=mx+b$ , where $m$ is the slope and $b$ is the $y$ -coordinate of the $y$ -intercept.
Standard form.: The equation $Ax+By=C$ . Solve for $y$ to read the slope, or use $m=-\frac{A}{B}$ when $B\neq 0$ .
Perpendicular bisector.: The line through the midpoint of a segment that is perpendicular to it; built from the midpoint and the opposite reciprocal slope.

Worked example

Line

k

passes through

(-2,7)

and

(4,-5)

. (a) Decide whether line

k

is parallel, perpendicular, or neither to the line

x-2y=10

. (b) Write, in slope-intercept form, the equation of the line through

(6,1)

that is perpendicular to line

k

Part (a), Step 1 — slope of line $k$ . Use the slope formula with

(-2,7)

as point 1 and

(4,-5)

as point 2:

m_k=\frac{-5-7}{4-(-2)}=\frac{-12}{6}=-2.

Step 2 — slope of $x-2y=10$ . Solve for

y

-2y=-x+10

, so

y=\frac{1}{2}x-5

. Its slope is

\frac{1}{2}

.

Step 3 — compare. The slopes

-2

and

\frac{1}{2}

are not equal, so the lines are not parallel. Multiply:

-2\cdot\frac{1}{2}=-1

. The product is

-1

, so line

k

is perpendicular to

x-2y=10

.

Part (b), Step 1 — the needed slope. A line perpendicular to

k

has the opposite reciprocal of

-2

. Flip

-\frac{2}{1}

to get

-\frac{1}{2}

, then change the sign:

m=\frac{1}{2}

. (Check:

-2\cdot\frac{1}{2}=-1

.)

Step 2 — point-slope form. With

(x_1,y_1)=(6,1)

y-1=\frac{1}{2}(x-6).

Step 3 — simplify. Distribute:

y-1=\frac{1}{2}x-3

. Add

1

y=\frac{1}{2}x-2

.

Step 4 — check. Substitute

x=6

y=\frac{1}{2}(6)-2=3-2=1

. The point

(6,1)

is on the line, and the slope

\frac{1}{2}

is the opposite reciprocal of

-2

. Both conditions hold.

Notice that this new line happens to be parallel to

x-2y=10

, which makes sense: two lines perpendicular to the same line are parallel to each other.

Practice questions

Which line is perpendicular to

3x+5y=20

$y=-\frac{3}{5}x+4$
$y=\frac{5}{3}x-1$
$y=-\frac{5}{3}x+2$
$y=\frac{3}{5}x+7$

Answer: $y=\frac{5}{3}x-1$

First find the slope of

3x+5y=20

. Solving for

y

5y=-3x+20

, so

y=-\frac{3}{5}x+4

, giving

m=-\frac{3}{5}

. The perpendicular slope is the opposite reciprocal: flip to

-\frac{5}{3}

, change the sign to

\frac{5}{3}

. Only

y=\frac{5}{3}x-1

has that slope, and

-\frac{3}{5}\cdot\frac{5}{3}=-1

confirms it. The choice

y=-\frac{3}{5}x+4

is the original line rewritten (parallel, in fact identical),

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

flips the fraction but forgets the sign change, and

y=\frac{3}{5}x+7

changes only the sign.

Triangle

PQR

has vertices

P(-3,1)

Q(1,9)

, and

R(5,3)

. Write the equation of the altitude from

R

to side

\overline{PQ}

, and explain why your slope is correct.

Answer: $y-3=-\frac{1}{2}(x-5)$ , or equivalently $y=-\frac{1}{2}x+\frac{11}{2}$ .

An altitude from

R

must be perpendicular to

\overline{PQ}

and pass through

R

. Slope of

\overline{PQ}

\frac{9-1}{1-(-3)}=\frac{8}{4}=2

. The altitude's slope is the opposite reciprocal of

2

, which is

-\frac{1}{2}

; checking,

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

. Now use point-slope with

R(5,3)

y-3=-\frac{1}{2}(x-5)

. Distributing gives

y-3=-\frac{1}{2}x+\frac{5}{2}

, so

y=-\frac{1}{2}x+\frac{11}{2}

. A frequent mistake is using the slope of

\overline{PR}

\overline{QR}

; the altitude is perpendicular to the side it drops onto, not to the sides it touches at the vertex.

Lines

a

and

b

are graphed. Line

a

contains

(0,-4)

and

(3,2)

. Line

b

contains

(-1,5)

and

(2,11)

. Are the lines parallel, perpendicular, the same line, or none of these? Justify.

Answer: Parallel.

Slope of

a

\frac{2-(-4)}{3-0}=\frac{6}{3}=2

. Slope of

b

\frac{11-5}{2-(-1)}=\frac{6}{3}=2

. Equal slopes mean the lines are parallel or identical, so check a point. Line

a

has equation

y=2x-4

. Testing

(-1,5)

from line

b

2(-1)-4=-6\neq 5

, so that point is not on line

a

. The lines are distinct with equal slopes, so they are parallel. Skipping the point check is the usual slip — equal slopes alone cannot rule out that the two descriptions name one single line.

FAQ

Why is the product of perpendicular slopes exactly $-1$ ?: Rotating a line $90^\circ$ about a point turns a rise of $a$ and a run of $b$ into a rise of $b$ and a run of $-a$ (or $-b$ and $a$ ). So the slope $\frac{a}{b}$ becomes $\frac{b}{-a}=-\frac{b}{a}$ , the opposite reciprocal. Multiplying the two gives $\frac{a}{b}\cdot\left(-\frac{b}{a}\right)=-1$ . The negative sign is what makes one line rise while the other falls.
What do I do when one line is vertical?: Do not try the multiplication test, because a vertical line has undefined slope and you cannot multiply by "undefined." Instead reason by form. A line perpendicular to the vertical line $x=c$ is horizontal, so through the point $(p,q)$ it is $y=q$ . A line parallel to $x=c$ through $(p,q)$ is $x=p$ . Similarly, perpendicular to $y=c$ through $(p,q)$ is $x=p$ .
Should I answer in point-slope or slope-intercept form?: Follow whatever the problem or your teacher asks for. Point-slope $y-y_1=m(x-x_1)$ is fastest to write because you just substitute. Slope-intercept $y=mx+b$ is better when you need to graph or compare $y$ -intercepts. They describe the same line, so converting is only a matter of distributing and adding.
Two lines have equal slopes but I got the same equation twice. Are they parallel?: No. If the slopes and the $y$ -intercepts both match, the two descriptions are of one single line, sometimes called coincident lines. Parallel requires the lines to be distinct and never intersect. Whenever slopes come out equal, test one point from the second line in the first line's equation; if it satisfies the equation, the lines coincide.

Learn this with a teacher, not a page

The Crimsora tutor teaches Slopes & Equations of Parallel and Perpendicular Lines live — explaining on a whiteboard, asking you questions, and adapting to where you get stuck.

Try a live session free