Two non-vertical lines are perpendicular exactly when the product of their slopes is negative one, so each slope is the negative reciprocal of the other. Includes testing a pair of lines, writing the equation of a line perpendicular to a given one through a given point, the horizontal-and-vertical special case, and why a graph can check but not prove perpendicularity.
Subject: Algebra 1 · 65 slides · symbolic lesson
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Title
Algebra 1 · Chapter 5 — Writing Linear Equations
Perpendicular Lines
Objectives
Five outcomes, each one you can test yourself on with a pencil and no answer key.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 5 Writing Linear Equations — Lesson 5.6 Perpendicular Lines §5.6, pp. 306-312 — the lesson these objectives are drawn from
Warm-up
Lesson 4.7 decided parallelism by comparing two slopes. This lesson decides the other important relationship by multiplying them.
Discussion prompt
The lines y equals 2x plus 3 and y equals negative one half x meet at a right angle. Multiply their two slopes and see what you get. Then try it on y equals 3x and y equals negative one third x.
Hint: Multiply, do not compare.
Answer:
\[ 2 \times \left(-\tfrac{1}{2}\right) = -1 \qquad 3 \times \left(-\tfrac{1}{3}\right) = -1 \]
Both products come to negative one. That is not a coincidence about these two pairs — it is the condition, and the Developing Concepts investigation on page 305 arrives at it by measuring angles.
Concept
Two lines in a plane are perpendicular if they intersect at a right angle. In a coordinate plane, two non-vertical lines are perpendicular if and only if the product of their slopes is negative one.
perpendicular — Two lines that intersect at a right angle. For non-vertical lines this happens exactly when the product of their slopes is negative one.
A proof of the relationship belongs to geometry; here it is used as a test.
Figure (svg): Two perpendicular lines with their slopes multiplied to give negative one
McDougal Littell Algebra 1: Concepts and Skills, Ch. 5 Writing Linear Equations — Lesson 5.6 Perpendicular Lines §5.6, pp. 305-306
Section
Section 1
Concept
The test is a single multiplication. Two non-vertical lines are perpendicular exactly when their slopes multiply to negative one, and not perpendicular otherwise.
\[ m_1 \cdot m_2 = -1 \]
The phrase if and only if means the test decides the question in both directions.
Figure (svg): Two perpendicular lines with their slopes multiplied to give negative one
McDougal Littell Algebra 1: Concepts and Skills, Ch. 5 Writing Linear Equations — Lesson 5.6 Perpendicular Lines §5.6, pp. 306-306 — the Perpendicular Lines box
Picture it
Two times negative a half.
Figure (svg): Two perpendicular lines with their slopes multiplied to give negative one
Nothing about the intercepts enters the test, so any two lines with these slopes are perpendicular wherever they happen to sit — which is what makes it a statement about direction alone.
Worked example
This is Example 1 from the textbook.
\[ \text{Are } \; y = -\tfrac{4}{3}x + 1 \; \text{ and } \; y = \tfrac{3}{4}x - 1 \; \text{ perpendicular?} \]
Read the two slopes
Why: Negative four thirds and three quarters.
\[ -\frac{4}{3}\text{ and } \frac{3}{4} \]
Multiply them
Why: The threes and fours cancel.
\[ -\frac{12}{12} \]
Simplify
Why: The product is negative one.
\[ -1 \]
Apply the condition
Why: A product of negative one means perpendicular.
Figure (svg): Two slopes multiplied to test a pair of lines
\[ -\tfrac{4}{3} \cdot \tfrac{3}{4} = -1 \]
Verify: notice what makes the cancellation work
Why: The numerator of one is the denominator of the other, so everything cancels except the sign. Any pair with that structure gives a product of negative one, which is a fast way to recognise perpendicular slopes without doing the multiplication.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 5 Writing Linear Equations — Lesson 5.6 Perpendicular Lines §5.6, pp. 306-306
Sorting
Multiply the two slopes each time.
Sort into buckets
Sort each pair of slopes by whether the lines are perpendicular.
The last item is the mirror-image error of the second: reciprocal sizes and matching signs give a product of positive one, which is just as wrong as negative four ninths.
Worked example
Guided Practice 2. The signs are opposite and the sizes are not.
\[ \text{Are } \; y = \tfrac{2}{3}x - 1 \; \text{ and } \; y = -\tfrac{2}{3}x + 1 \; \text{ perpendicular?} \]
Read the two slopes
Why: Two thirds and negative two thirds.
\[ \frac{2}{3}\text{ and } -\frac{2}{3} \]
Multiply them
Why: Four ninths, with a negative sign.
\[ -\frac{4}{9} \]
Compare with negative one
Why: Negative four ninths is not negative one.
\[ \text{not } -1 \]
Apply the condition
Why: The lines are not perpendicular.
Figure (svg): Two slopes multiplied to test a pair of lines
\[ \tfrac{2}{3} \cdot \left(-\tfrac{2}{3}\right) = -\tfrac{4}{9} \]
Verify: say what these two lines are instead
Why: They are mirror images of each other in a horizontal line, since one rises exactly as steeply as the other falls. That is a real relationship and it is not perpendicularity — the two lines meet at an angle noticeably wider than ninety degrees.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 5 Writing Linear Equations — Lesson 5.6 Perpendicular Lines §5.6, pp. 306-306
Trap
\[ m_1 = \tfrac{2}{3}, \quad m_2 = -\tfrac{2}{3} \]
Declare them perpendicular, since one rises and the other falls
Why: A right angle does involve one line going up and one down, so opposite signs feel like the condition.
The product is negative four ninths rather than negative one. Opposite signs are necessary and nowhere near sufficient.
\[ \tfrac{2}{3} \cdot \left(-\tfrac{3}{2}\right) = -1 \]
Multiply the two slopes and compare the product with negative one
Why: The sizes must be reciprocal as well as the signs opposite.
Doing the multiplication takes a few seconds and removes the guesswork entirely.
Faded example
Multiply and compare with negative one.
Fill in the blanks
-\tfrac12-1 \cdot \tfrac______ = -\dfrac______} = ___
Why: The numerators multiply to twelve and so do the denominators, so everything cancels except the sign. That structure — one fraction's numerator being the other's denominator — is exactly what perpendicular slopes look like.
Elimination
Multiply, do not compare.
Eliminate the wrong options
Which pair of slopes gives a product of -1?
Survives elimination: A
Why: Only the first has both properties: reciprocal sizes and opposite signs. Options B and C each get one of the two right, which is why the condition has to be checked by multiplying rather than by eye — each looks plausible on its own.
Socratic
The number is oddly specific for a geometric condition.
Discussion prompt
Give an informal reason why perpendicularity should involve flipping a slope and changing its sign, thinking about what happens to a rise-and-run triangle when you rotate it a quarter turn.
Hint: Rotate the little triangle and see where the rise and run end up.
Answer:
Take a line with slope three over four, so its triangle is four across and three up. Rotate that triangle a quarter turn and the four that pointed right now points up, while the three that pointed up now points left. So the new triangle is three across in the negative direction and four up, giving a slope of four over negative three.
That is exactly the negative reciprocal, and multiplying the two gives negative one because the numerator and denominator swap and one sign flips. The full proof belongs to geometry, as the textbook says, and this rotation argument is enough to make the rule stop feeling arbitrary.
Section
Section 2
Concept
Solving the condition for one slope shows that each is the negative reciprocal of the other. To find it, turn the fraction upside down and change its sign.
\[ m_1 m_2 = -1 \;\Longrightarrow\; m_2 = -\dfrac{1}{m_1} \]
A whole number has a hidden denominator of one, so its reciprocal is a fraction.
Figure (svg): A slope turned into its negative reciprocal in two steps
McDougal Littell Algebra 1: Concepts and Skills, Ch. 5 Writing Linear Equations — Lesson 5.6 Perpendicular Lines §5.6, pp. 306-306 — the Perpendicular Lines box, rearranged
Picture it
Flip, then negate.
Figure (svg): A slope turned into its negative reciprocal in two steps
Doing only one of the two moves is the standard error, and each half produces a recognisable wrong product: flipping alone gives positive one and negating alone gives minus the square.
Worked example
Whole numbers, fractions and a negative.
\[ \text{Find the negative reciprocal of } \; 3, \quad \tfrac{3}{4}, \quad -2, \quad -\tfrac{2}{5}. \]
Take 3
Why: Write it as three over one, flip to one third, negate.
\[ -\frac{1}{3} \]
Take three quarters
Why: Flip to four thirds, negate.
\[ -\frac{4}{3} \]
Take negative 2
Why: Flip to negative one half, negate.
\[ \frac{1}{2} \]
Take negative two fifths
Why: Flip to negative five halves, negate.
\[ \frac{5}{2} \]
Figure (svg): A slope turned into its negative reciprocal in two steps
\[ -\tfrac{1}{3}, \quad -\tfrac{4}{3}, \quad \tfrac{1}{2}, \quad \tfrac{5}{2} \]
Verify: multiply each pair
Why: Three times negative a third is negative one, and so are the other three products. Checking by multiplication is faster than rechecking the two moves, and it tests both of them at once.
Translation
Flip, then change the sign.
Match the pairs
Why: A whole number has a hidden denominator of one, so its negative reciprocal is a fraction with one on top. The last two start negative and end positive, since changing the sign of a negative gives a positive — the rule applies in both directions.
Worked example
The slope has to be extracted before it can be flipped.
\[ \text{What slope is perpendicular to the line } \; 4x + 2y = 10? \]
Rewrite in slope-intercept form
Why: Isolating y gives y equals negative 2x plus five.
\[ m = -2 \]
Flip the slope
Why: Negative two is negative two over one, so flipping gives negative one half.
\[ -\frac{1}{2} \]
Change the sign
Why: The negative reciprocal is one half.
\[ \frac{1}{2} \]
Check
Why: Negative two times a half is negative one.
Figure (svg): The solution to Worked example read a perpendicular slope off an equation shown as a ladder of expressions, one row per algebraic move
\[ m = -2 \;\Longrightarrow\; m_\perp = \tfrac{1}{2} \]
Verify: notice why the rewriting was necessary
Why: Read straight off the standard form, the coefficient of x is four, which would have given a perpendicular slope of negative one quarter — wrong. Only the slope of a line in slope-intercept form can be flipped, which is the same warning Lesson 4.7 gave about parallelism.
Error analysis
The student was asked for the slope perpendicular to three quarters.
Annotate
On: \( \begin{aligned} m &= \tfrac{3}{4} \\ m_\perp &= -\tfrac{3}{4} \\ \text{check: } \tfrac{3}{4} \cdot \left(-\tfrac{3}{4}\right) &= -\tfrac{9}{16} \end{aligned} \)
Negating alone always produces minus the square of the slope, which equals negative one only when the slope is one or negative one. Recognising that signature identifies this error instantly.
Faded example
Both moves, in either order.
Fill in the blanks
m = \tfrac4-: \quad \text___ \tfrac___}___, \; \text___ ___\tfrac______
Why: Flipping exchanges the numerator and denominator, and negating changes the sign, giving negative four thirds. Multiplying three quarters by it gives negative twelve twelfths, which is negative one, confirming both moves were made.
Elimination
Asked for the slope perpendicular to 2, a student answers -4.
Eliminate the wrong options
What most likely went wrong?
Survives elimination: A
Why: The correct answer is negative one half, and negative four is the result of doubling rather than reciprocating. Each of the standard errors has its own signature — minus the original, the plain reciprocal, or minus a multiple — and recognising which one you are looking at is faster than rechecking every step.
Socratic
Most slopes are not.
Discussion prompt
Find every slope that is its own negative reciprocal, and say what that means about the pair of lines involved. Then say what it would mean for a slope to be its own reciprocal without the sign change.
Hint: Set the slope equal to its own negative reciprocal and solve.
Answer:
Setting m equal to negative one over m gives m squared equal to negative one, which no real number satisfies. So no slope is its own negative reciprocal, meaning no line is perpendicular to another line with the same slope — which makes sense, since lines with equal slopes are parallel and never meet at all.
A slope that is its own reciprocal satisfies m squared equal to one, so m is one or negative one. Those are the two lines at forty-five degrees to the axes, and they are perpendicular to each other rather than to themselves: one times negative one is negative one. That pair is the only one where the two slopes have the same size.
Section
Section 3
Concept
To write the equation of a line perpendicular to a given one through a given point, take the negative reciprocal of the given slope and use point-slope form with the point.
The route is Lesson 5.2's with one extra step at the front.
Figure (svg): A perpendicular line constructed through a given point
McDougal Littell Algebra 1: Concepts and Skills, Ch. 5 Writing Linear Equations — Lesson 5.6 Perpendicular Lines §5.6, pp. 307-307 — Example 2, Show that Lines are Perpendicular
Picture it
The slope is transformed; the point is used as given.
Figure (svg): A perpendicular line constructed through a given point
Compare this with the parallel construction in Lesson 5.2. The only difference is that the slope is transformed rather than copied, and everything after that step is identical.
Worked example
This is Example 2 from the textbook, both parts.
\[ \text{Write the line through } (2, 5) \text{ and } (4, 4), \text{ then show it is perpendicular to } y = 2x - 1. \]
Compute the slope from the two points
Why: Four minus five over four minus two.
\[ m = -\frac{1}{2} \]
Write point-slope form
Why: Using (2, 5).
\[ y - 5 = -(\frac{1}{2}) (x - 2) \]
Distribute and isolate y
Why: y minus five equals negative one half x plus one.
\[ y = -(\frac{1}{2}) x + 6 \]
Multiply the two slopes
Why: Negative one half times two.
\[ -1,\text{ perpendicular} \]
Figure (svg): A perpendicular line constructed through a given point
\[ y = -\tfrac{1}{2}x + 6 \qquad -\tfrac{1}{2} \cdot 2 = -1 \]
Verify: check both given points in the equation
Why: At x equal to two the equation gives negative one plus six, which is five, and at four it gives negative two plus six, which is four. Both points check, and the slope product settles the perpendicularity — two separate claims needing two separate checks.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 5 Writing Linear Equations — Lesson 5.6 Perpendicular Lines §5.6, pp. 307-307
Faded example
Transform the slope, then use the point.
Fill in the blanks
\perp \text- y = 2x - 1 \text2 (2, 5): \quad m_\perp = ___\tfrac______, \; y - 5 = -\tfrac______(x - ___)
Why: The given slope of two flips to one half and negates to negative one half. Distributing and isolating y gives y equals negative one half x plus six, and multiplying the two slopes gives negative one, confirming the perpendicularity.
Worked example
Guided Practice 3 and 4. Two points each, and a line to compare with.
\[ \text{Through } (1, 3) \text{ and } (3, 6), \text{ against } y = -\tfrac{2}{3}x + 5; \text{ and through } (0, 0) \text{ and } (1, 2), \text{ against } y = -\tfrac{1}{2}x + 7. \]
First pair: compute the slope
Why: Six minus three over three minus one.
\[ m = \frac{3}{2} \]
Test against the given line
Why: Three halves times negative two thirds is negative one.
Second pair: compute the slope
Why: Two minus zero over one minus zero.
\[ m = 2 \]
Test against the given line
Why: Two times negative one half is negative one.
Figure (svg): The solution to Worked example two from guided practice shown as a ladder of expressions, one row per algebraic move
\[ \tfrac{3}{2} \cdot \left(-\tfrac{2}{3}\right) = -1 \qquad 2 \cdot \left(-\tfrac{1}{2}\right) = -1 \]
Verify: notice the second line's intercept
Why: It passes through the origin, so it is a direct variation model from Lesson 4.6 — and being perpendicular to something has nothing to do with where a line sits, only with its direction. Two lines with these slopes are perpendicular wherever they are placed.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 5 Writing Linear Equations — Lesson 5.6 Perpendicular Lines §5.6, pp. 307-307
Trap
\[ \text{perpendicular to } y = 2x - 1 \text{ through } (2, 5) \]
Use slope 2 and write y - 5 = 2(x - 2)
Why: The parallel construction from Lesson 5.2 copied the slope, and the two problems look identical.
That gives a line parallel to the given one rather than perpendicular. Its slope product is four rather than negative one.
\[ m_\perp = -\tfrac{1}{2} \;\Longrightarrow\; y - 5 = -\tfrac{1}{2}(x - 2) \]
Transform the slope first, then use the point exactly as before
Why: Parallel copies the slope; perpendicular flips and negates it.
Multiplying the two slopes at the end is the check, and it distinguishes the two constructions immediately.
Comparison
Fill the blanks from memory before you scroll back.
Comparison matrix
| Parallel through a point | Perpendicular through a point | |
|---|---|---|
| What happens to the slope | copied unchanged | flipped and negated |
| What happens to the point | used as given | used as given |
| The check at the end | slopes equal, intercepts differ | the product of the slopes is -1 |
Only the first row differs. Everything else about the two constructions is the same, which is why they are so easy to confuse and why the final check matters.
Elimination
Perpendicular to y equals 3x plus 1, through (6, 2).
Eliminate the wrong options
Which equation is right?
Survives elimination: A
Why: The negative reciprocal of three is negative one third, and point-slope through (6, 2) gives y equals negative one third x plus four. Options B and C each satisfy one of the two requirements, which is why both have to be checked: the slope product for perpendicularity, and a substitution for the point.
Socratic
The given line has infinitely many perpendiculars.
Discussion prompt
Explain how many lines are perpendicular to a given line, and why specifying a point picks out exactly one of them. Then say what that count has in common with Lesson 5.2's parallel problem.
Hint: Ask what the perpendicular condition fixes.
Answer:
The condition fixes only the slope, and every line with that slope is perpendicular to the given one — infinitely many parallel lines, all at right angles to the original. Specifying a point then picks the single one of them that passes through it, since one point and a slope determine one line.
The parallel problem has exactly the same structure: the condition fixes the slope and the point selects one line from the infinite family. Both constructions are two facts assembled by point-slope form, and they differ only in how the slope is obtained — which is why they should be practised together rather than separately.
Section
Section 4
Concept
A horizontal line and a vertical line are perpendicular. The product rule cannot say so, because a vertical line has no slope to multiply, which is why the rule is stated for non-vertical lines and this case declared separately.
The axes themselves are the simplest example.
Figure (svg): A horizontal and a vertical line meeting at a right angle
McDougal Littell Algebra 1: Concepts and Skills, Ch. 5 Writing Linear Equations — Lesson 5.6 Perpendicular Lines §5.6, pp. 306-306 — the note that horizontal and vertical lines are perpendicular to each other
Picture it
Slope zero, and no slope.
Figure (svg): A horizontal and a vertical line meeting at a right angle
The vertical line has been the exception in every lesson since Chapter 4, and always for the same reason: its run is zero, so it has no slope for any rule about slopes to use.
Worked example
Attempting the product shows what goes wrong.
\[ \text{Try to apply the product rule to } \; y = 1 \; \text{ and } \; x = 2. \]
Find the first slope
Why: A horizontal line has slope zero.
\[ m = 0 \]
Find the second slope
Why: A vertical line's run is zero, so it has no slope.
Try to multiply
Why: There is no second number to multiply by.
Note the resolution
Why: The lines are perpendicular by a separate statement rather than by the rule.
Figure (svg): A horizontal and a vertical line meeting at a right angle
\[ y = 1 \perp x = 2 \text{, by the separate statement} \]
Verify: ask what would happen if you pretended the slope were zero
Why: Zero times zero is zero rather than negative one, so the rule would call them not perpendicular — which contradicts the picture plainly. That contradiction is why the textbook has to state the case separately rather than letting the rule handle it.
Sorting
Watch for the special case.
Sort into buckets
Sort each pair of lines by whether they are perpendicular.
Every pair in the second bucket is parallel, which is the opposite relationship. Parallel and perpendicular are the two things a pair of lines can be, and everything else is neither.
Worked example
The construction, in the case where the answer has no slope.
\[ \text{Write the line perpendicular to } y = 4 \text{ through } (3, -1). \]
Identify the given line
Why: It is horizontal, with slope zero.
Try the negative reciprocal
Why: One divided by zero has no value.
Use the special case instead
Why: A perpendicular to a horizontal line is vertical.
Use the point
Why: The vertical line through (3, -1) has x equal to three.
\[ x = 3 \]
Figure (svg): A horizontal and a vertical line meeting at a right angle
\[ x = 3 \]
Verify: check the point and the direction
Why: The point (3, -1) has an x-coordinate of three, so it is on the line, and a vertical line does meet a horizontal one at a right angle. Neither slope-intercept nor point-slope form could have written this answer, which is why Lesson 5.4's standard form matters.
Trap
\[ \text{perpendicular to } y = 4: \; m = 0 \]
Take the negative reciprocal: -1/0 = 0
Why: The procedure is applied without noticing that the slope is zero.
One divided by zero has no value, so there is no negative reciprocal to take. Reporting zero would give a second horizontal line, parallel to the first rather than perpendicular.
A perpendicular to a horizontal line is vertical, so the answer is x equals a constant.
Check whether the given slope is zero before reciprocating
Why: Zero is the one slope with no reciprocal, and it is exactly the case the special statement covers.
The same warning applies in reverse: a line perpendicular to a vertical one is horizontal, and the vertical line had no slope to start from.
Faded example
The answer has no slope.
Fill in the blanks
A line perpendicular to y = 4 is vertical, and the one through (3, -1) is x = 3.
Why: A vertical line is perpendicular to every horizontal one, and the point's x-coordinate fixes which vertical line it is. No slope is computed anywhere, because neither the given line's slope of zero nor the answer's non-existent slope can be used in the product rule.
Prediction
The given line is vertical.
Predict first
Which line is perpendicular to x = 5?
Correct: Any horizontal line, such as y = 2.
\[ x = 5 \perp y = b \text{ for every } b \]
Why: A vertical line meets every horizontal line at a right angle, which the separate statement covers. The third option treats the five as a slope, which it is not — it is an x-coordinate. And the fourth mistakes a limitation of the product rule for a fact about geometry: the lines are perpendicular whether or not the rule can say so.
Socratic
It would be tidier if one rule covered everything.
Discussion prompt
Explain why the product rule cannot be extended to cover vertical lines, even with a convention about what a vertical slope means. Then say what other rules in Chapters 4 and 5 had to make the same exception.
Hint: Ask what number a vertical line's slope would have to be.
Answer:
A vertical line's slope would have to be a number that multiplies zero to give negative one, and no such number exists — zero times anything is zero. So no convention can rescue the rule; the failure is not about which value to pick but about there being none that works.
The same exception appears throughout: slope-intercept form and point-slope form cannot write a vertical line, the vertical line test excludes it from being a function, the slope formula divides by zero for it, and Lesson 4.4 found it has no y-intercept. All five failures come from the single fact that its run is zero, which is worth more than remembering five exceptions.
Section
Section 5
Concept
Graphing can show that an answer is reasonable. It cannot show that two lines are perpendicular, because a drawing cannot distinguish a right angle from one a degree or two away.
The textbook states this restriction directly on page 307.
Figure (svg): Two nearly perpendicular lines that a drawing cannot distinguish
McDougal Littell Algebra 1: Concepts and Skills, Ch. 5 Writing Linear Equations — Lesson 5.6 Perpendicular Lines §5.6, pp. 307-307 — the note that graphing cannot be used to show that two lines are perpendicular
Picture it
Only one of these is perpendicular.
Figure (svg): Two nearly perpendicular lines that a drawing cannot distinguish
The right-hand pair meets at about ninety-three degrees, which no drawn figure distinguishes from ninety. Every judgement made from the picture alone would get both of these wrong half the time.
Worked example
Two pairs of slopes that look identical on a page.
\[ \text{Compare the pairs } \; (2, -\tfrac{1}{2}) \; \text{ and } \; (2, -0.45). \]
Multiply the first pair
Why: Two times negative a half.
\[ -1 \]
Multiply the second pair
Why: Two times negative 0.45.
\[ -0.9 \]
Compare the verdicts
Why: Only the first is perpendicular.
Consider the drawings
Why: The two second lines differ by about three degrees.
Figure (svg): Two nearly perpendicular lines that a drawing cannot distinguish
\[ 2 \cdot \left(-\tfrac{1}{2}\right) = -1 \qquad 2 \cdot (-0.45) = -0.9 \]
Verify: ask what precision a drawing offers
Why: A line drawn by hand on a grid is accurate to perhaps two or three degrees at best, which is exactly the size of the difference here. So the picture cannot resolve the question, while one multiplication does so exactly.
Elimination
Two lines are drawn and appear to meet at a right angle.
Eliminate the wrong options
What may you conclude?
Survives elimination: A
Why: The sketch is a filter rather than a proof: it rejects obviously wrong answers quickly and cannot confirm a correct one. Option D overcorrects, and knowing what a coarse check is genuinely good for is as useful as knowing its limits.
Worked example
The picture still earns its place.
\[ \text{A student answers } y = 2x + 6 \text{ for a line perpendicular to } y = 2x - 1. \text{ What does a sketch show?} \]
Sketch both lines
Why: They rise at the same rate.
Judge by eye
Why: They never meet, let alone at a right angle.
Confirm algebraically
Why: Two times two is four, not negative one.
Note the division of labour
Why: The sketch caught it instantly; the algebra proved it.
Figure (svg): Two nearly perpendicular lines that a drawing cannot distinguish
\[ 2 \cdot 2 = 4 \neq -1 \]
Verify: classify what each method caught
Why: The sketch catches gross errors — a copied slope, a wrong sign, a line through the wrong region — because those are visible at a glance. It cannot catch a slope that is slightly off, which is where the product rule takes over. Using each for what it is good at is faster than using either alone.
Trap
Two lines are graphed and appear to meet at a right angle.
Report that they are perpendicular
Why: The picture is convincing and the angle looks square.
A drawing cannot distinguish ninety degrees from eighty-seven. The pair might have a slope product of negative nine tenths, which no sketch will reveal.
\[ \text{compute } m_1 m_2 \text{ and compare with } -1 \]
Use the sketch to check reasonableness and the product to decide
Why: The textbook says this explicitly: graphing checks, it does not show.
The same limitation applies to reading a slope off a graph, which is why Lesson 5.1 insisted on exact grid crossings.
Faded example
The pictures look alike; the products do not.
Fill in the blanks
2 \cdot \left(-\tfrac-1-0.9\right) = ___ \qquad 2 \cdot (-0.45) = ___
Why: Only the first product is negative one, so only the first pair is perpendicular. The second is off by a tenth, which corresponds to about three degrees — well inside the accuracy of any drawing and well outside what the condition allows.
Two truths and a lie
Three statements about perpendicular lines. Two are true and one is not.
Eliminate the wrong options
Which statement is false?
Survives elimination: A
Why: The first is false, and it is the most tempting of the four. A drawing cannot resolve the difference between a product of negative one and negative nine tenths, which is about three degrees, so appearance is evidence rather than proof. The other three follow directly from the condition being a statement about slopes alone.
Socratic
The condition is stated with that phrase deliberately.
Discussion prompt
Explain what the phrase if and only if means in the perpendicular condition, and what would be lost if it said only if. Then give a question each direction of the statement answers.
Hint: The statement can be read in two directions.
Answer:
It means the implication runs both ways: perpendicular lines have a slope product of negative one, and any two lines with a slope product of negative one are perpendicular. With only if, you could conclude the product from perpendicularity but not perpendicularity from the product — so the test would be able to reject a pair and never confirm one.
The first direction answers questions like: these lines meet at a right angle, so what must their slopes satisfy? The second answers: I computed a product of negative one, so what do I now know? Almost every use in this lesson is the second direction, which is exactly the half a one-way statement would not give you.
Comparison
Fill the blanks from memory before you scroll back.
Comparison matrix
| Parallel | Perpendicular | |
|---|---|---|
| Condition on slopes | equal | product is -1 |
| The lines | never meet | meet at a right angle |
| The special case | any two vertical lines | one horizontal and one vertical |
Both relationships are decided by the slopes alone and both need a separate statement for vertical lines. The difference is comparison against multiplication.
Pattern
Whether you are testing a pair or constructing one, the same five moves cover it.
Step one takes a glance and prevents the attempt to take a reciprocal of zero, which is the one case the whole rule cannot handle.
OpenStax Elementary Algebra 2e, §4.6 Find the Equation of a Line §4.6
Check
Multiply, do not compare.
Check your understanding
Which pair of slopes belongs to perpendicular lines?
Answer: A
Why: Four times negative one quarter is negative one, so the sizes are reciprocal and the signs opposite. Both properties are needed and only this pair has both.
Check
Flip, then negate.
Check your understanding
What slope is perpendicular to -2/3?
Answer: A
Why: Flipping negative two thirds gives negative three halves, and changing the sign gives three halves. Multiplying negative two thirds by three halves gives negative one, confirming it.
Check
The special case.
Check your understanding
Which line is perpendicular to y = -3?
Answer: A
Why: A horizontal line and a vertical line are perpendicular, which the textbook states separately from the product rule. Zero has no reciprocal, so the rule cannot handle this case, and the lines are perpendicular regardless.
Real world
This is Example 3's situation. A helicopter is flying along a path and needs to reach a straight highway by the shortest possible route.
Discussion prompt
Explain why the shortest path from the helicopter to the highway is perpendicular to it, and describe how you would find that path's equation given the highway's equation and the helicopter's position.
Hint: Think about what the shortest distance from a point to a line looks like.
Answer:
The shortest distance from a point to a line is measured along the perpendicular. Any other route reaches the line at a slant, and the slanted segment is the hypotenuse of a right triangle whose shorter leg is the perpendicular one — so every other path is longer.
\[ \text{highway } y = mx + b, \; \text{helicopter at } (x_1, y_1) \;\Longrightarrow\; y - y_1 = -\tfrac{1}{m}(x - x_1) \]
So the procedure is: read the highway's slope, take its negative reciprocal, and use point-slope form with the helicopter's position. Where the two lines cross is the landing point, and finding that crossing is a system of two equations — which is exactly what Chapter 7 is about.
Commit first
Answer, then rate your confidence honestly. Confident and wrong is the combination worth finding.
Predict first
Are the lines y = (2/3)x - 1 and y = -(2/3)x + 1 perpendicular?
Correct: No, since the product of the slopes is -4/9 rather than -1.
\[ \tfrac{2}{3} \cdot \left(-\tfrac{2}{3}\right) = -\tfrac{4}{9} \neq -1 \]
\[ \text{for a perpendicular: } \tfrac{2}{3} \cdot \left(-\tfrac{3}{2}\right) = -1 \]
Why: Opposite signs are necessary and not sufficient — the two sizes must also be reciprocal, and two thirds is not the reciprocal of two thirds. The product is negative four ninths, so the lines meet at an angle noticeably wider than ninety degrees. This is Guided Practice 2 in the textbook, and the pair is designed to look perpendicular: the two lines are mirror images of each other, which is a real relationship and a different one.
Explain it
They know parallel lines have equal slopes and think perpendicular ones have opposite slopes.
Discussion prompt
In no more than four sentences, correct that idea and give them a reliable test. Then tell them the one case where the test does not apply.
Hint: Two moves, not one.
Answer:
A usable answer: opposite signs are only half of it — the sizes have to be reciprocal too, so the perpendicular to a slope of two thirds is negative three halves rather than negative two thirds. The test is to multiply the two slopes: if you get exactly negative one they are perpendicular, and anything else means they are not.
The case where it does not apply is a vertical line, which has no slope at all, so there is nothing to multiply. A vertical line is perpendicular to every horizontal one, and that has to be remembered separately rather than computed.
Exit ticket
Name the weakest spot before you close the deck. That is the one worth ten minutes tonight.
Predict first
Which of these would you least want to be handed cold on a quiz tomorrow?
Correct: Whichever you picked is the right answer — and each one has a specific fix.
Why: The two moves are fixed by multiplying your answer by the original and checking you get negative one. Transform-versus-copy is fixed by asking whether the question said parallel or perpendicular before touching the slope. The special case is fixed by glancing at the given line first, since a slope of zero has no reciprocal. Judging from a drawing is fixed by remembering that three degrees is invisible and changes the answer. Pick yours and do five of that kind tonight rather than twenty mixed ones.
Connect it up
Do this on paper. It is worth more than rereading the slides.
Draw it
At the top of a page write four slopes, including a whole number, a fraction and a negative, and beside each write its negative reciprocal and the product of the two, checking that every product is negative one. Underneath, choose one of your slopes, write a line with it, and construct the perpendicular through a point of your choosing, showing the negative reciprocal step separately from the point-slope step. To the right, draw a coordinate plane and graph both lines, marking the right angle where they cross and writing beside it that the drawing is a check rather than a proof. In the lower half, draw a horizontal line and a vertical line, write both equations, and write one sentence explaining why the product rule cannot be used on them. Finally, in the margin, write down what a slope product of positive one, of negative four ninths, and of negative one each tell you about a pair of lines.
Every product in your top row should be exactly negative one. A product of positive one means you flipped and forgot to negate, and a product that is minus a square means you negated and forgot to flip — each wrong answer names its own error.
Recap
Five things, and the first is a multiplication rather than a comparison.
| If the question says | Your first move is |
|---|---|
| Are these lines perpendicular | Read both slopes and multiply |
| Perpendicular to this line, through this point | Flip and negate the slope |
| The given line is horizontal | The answer is vertical: x = a constant |
| They look perpendicular on the graph | Check the slope product before agreeing |
| Find the shortest path to the line | Build the perpendicular through the point |
That completes Chapter 5. Chapter 6 replaces the equals sign with an inequality: instead of the points on a line, the solutions become whole regions of the plane, and the techniques of Chapter 3 have to be adapted to handle them.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 5 Writing Linear Equations — Lesson 5.6 Perpendicular Lines §5.6, pp. 306-312 — everything on these slides traces back here
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