Graphs lines quickly from the intercept and the slope rather than from a table, finds both intercepts, and establishes the two relationships between pairs of lines: parallel means equal slopes, and perpendicular means slopes whose product is negative one. Closes by finding where two lines cross, which is a two-variable system solved before the machinery for it arrives in Chapter 9.
Subject: Precalculus · 65 slides · symbolic lesson
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Title
Precalculus · Chapter 2 — Linear Functions
§2.2 Graphs of Linear Functions, pp. 205-232
Objectives
Five things, each one you can check yourself on paper.
OpenStax, Precalculus, §2.2 Graphs of Linear Functions §2.2, pp. 205-232 — the pages these objectives are drawn from
Warm-up
Graphing by table is safe and slow. This section is about doing it in two marks on the page.
Discussion prompt
How many points determine a line, and why is plotting exactly that many nevertheless a bad idea?
Hint: What do two points always look like, whether or not they are right?
Answer:
Two points determine a line, so two is enough in principle and there is no need for a table of five.
The trouble is that any two points look collinear. If you make an arithmetic slip in one of them, the line you draw through them is wrong and looks perfectly convincing — there is nothing to disagree with.
So the practical answer is three: two to determine it and one to check. If the third is off the line, one of the three is wrong and you know to look. That costs almost nothing and catches the errors that a table of five was really there to catch.
Concept
A line is graphed from two pieces of information: a point to begin at, and a direction to move in. The slope-intercept form supplies exactly those two.
\[ y=mx+b: \quad \text{start at } (0,b), \text{ then move } \Delta x = 1, \; \Delta y = m \]
Reading the slope as a pair of instructions — over by the run, then up by the rise — turns graphing into a staircase rather than a computation. A negative slope steps down instead of up, and a fractional slope is easier walked as its actual rise and run than as a decimal.
Figure (svg): A line graphed by plotting its y-intercept and then stepping repeatedly by the rise over the run, with each step drawn as a small right-angled staircase
OpenStax, Precalculus, §2.2 Graphs of Linear Functions §2.2, pp. 205-210
Section
Section 1
Concept
Start at the y-intercept and step by the slope, treating it as a rise over a run. Two or three points drawn this way determine the line.
A whole-number slope is a fraction over 1, so a slope of 3 means right 1 and up 3. A slope written as a decimal is worth converting: stepping by 0.6 is awkward, while stepping right 5 and up 3 lands exactly on a lattice point and is easier to draw accurately.
Figure (svg): A line graphed by plotting its y-intercept and then stepping repeatedly by the rise over the run, with each step drawn as a small right-angled staircase
OpenStax, Precalculus, §2.2 Graphs of Linear Functions §2.2, pp. 205-212
Picture it
Each step is one run across and one rise up, repeated.
Figure (svg): A line graphed by plotting its y-intercept and then stepping repeatedly by the rise over the run, with each step drawn as a small right-angled staircase
The steps all have the same shape because the slope is constant, which is §2.1's definition drawn as a procedure.
Worked example
Convert the slope into a step before drawing anything.
\[ \text{Graph } y = \tfrac{2}{3}x + 1. \]
Plot the intercept
Why: The output at the input zero.
\[ \text{point } (0, 1) \]
Read the slope as a step
Why: Numerator is rise, denominator is run.
\[ \text{right } 3,\text{ up } 2 \]
Step once from the intercept
Why: Landing on a lattice point.
\[ \text{point } (3, 3) \]
Step again as a check
Why: A third point must be collinear.
\[ \text{point } (6, 5) \]
Figure (svg): The solution to Worked example graph from slope-intercept form shown as a ladder of expressions, one row per legal move
\[ \text{through } (0,1), \, (3,3), \, (6,5) \]
Verify: confirm the third point from the rule
Why: Substituting 6 gives two thirds of 6, which is 4, plus 1, giving 5. The stepped point and the computed point agree, so the staircase was walked correctly. Had they disagreed, the error would be in one step and easy to find.
OpenStax, Precalculus, §2.2 Graphs of Linear Functions §2.2, pp. 206-208
Matching
Read the numerator as the rise and the denominator as the run.
Match the pairs
Why: A whole number is a fraction over 1, so its run is 1. A negative slope keeps the rightward run and reverses the vertical move. Stepping by the actual numerator and denominator rather than by a decimal lands on lattice points, which is what makes a hand-drawn line accurate.
Worked example
The run still goes right; the rise goes down.
\[ \text{Graph } y = -\tfrac{3}{4}x + 5. \]
Plot the intercept
Why: Height 5 on the vertical axis.
\[ \text{point } (0, 5) \]
Read the step, keeping the sign
Why: Right 4, and down 3 because the slope is negative.
\[ \text{right } 4,\text{ down } 3 \]
Step once
Why: From the intercept.
\[ \text{point } (4, 2) \]
Step again
Why: Continuing in the same direction.
\[ \text{point } (8, -1) \]
Figure (svg): The solution to Worked example a negative slope shown as a ladder of expressions, one row per legal move
\[ \text{through } (0,5), \, (4,2), \, (8,-1) \]
Verify: check the direction against the sign
Why: The line should fall as it goes right, and each stepped point is lower than the last — consistent with a negative slope. Attaching the minus sign to the run instead, stepping left 4 and up 3, gives points on the same line, which is worth noticing: it lands on the line's other side and is equally valid.
OpenStax, Precalculus, §2.2 Graphs of Linear Functions §2.2, pp. 208-210
Trap
\[ y = -2x + 3: \quad \text{from } (0,3), \text{ step right } 1 \text{ and up } 2 \]
Read the slope's size and step by it
Why: The number 2 is used as the rise and the sign is dropped.
The stepped point is taken as (1, 5), and the line is drawn rising.
The sign belongs to the step. A slope of negative 2 means right 1 and DOWN 2, so the next point is at (1, 1) and the line falls.
Check against the rule: at the input 1 it gives negative 2 plus 3, which is 1 — not 5.
Before stepping, say aloud whether the line rises or falls. A negative slope falls to the right, and one glance at the finished sketch will then confirm or contradict it.
Prediction
A line passes through the point at (1, 2) and has slope 4.
Predict first
Which point is also on it?
Correct: (2, 6).
Why: Stepping right 1 from the input 1 raises the output by the slope, 4, giving 2 plus 4, which is 6. The second option steps by the slope in the wrong direction, treating 4 as a run. The third steps by 1 vertically, ignoring the slope's size altogether.
Faded example
A line has intercept negative 3 and slope five halves. Find the next lattice point.
Fill in the blanks
\text2 (0,-3) \;\Longrightarrow\; \text5 ___, \text___ ___ \;\Longrightarrow\; (2,\,2)
Why: The denominator 2 is the run and the numerator 5 is the rise, so the step is right 2 and up 5, landing at the point with coordinates 2 and 2. Checking: the rule at the input 2 gives 5 minus 3, which is 2.
Step zero
You are asked to graph a line given in the form with x and y both on the left.
Discussion prompt
What must you do before you can step off the slope, and is there a faster route?
Hint: Which form displays the intercept and slope, and which two points are easiest to find?
Answer:
One route is to solve for y, putting it into slope-intercept form so that the intercept and slope are visible, then step as usual.
The faster route for that form is often both intercepts. Setting x to zero and then y to zero each takes one step and gives a point on an axis, and two points determine the line without any stepping at all.
The intercept route fails only when the line passes through the origin, since then both intercepts are the same point. In that case one extra point has to be computed, which is why the rearranging route is the one that always works.
Section
Section 2
Concept
An intercept is where a graph meets an axis. On each axis the other coordinate is zero, so setting that coordinate to zero and solving finds the intercept.
The pairing is the thing to keep straight, and it is genuinely counterintuitive: the y-intercept is found by setting x to zero. Saying which variable you are setting to zero out loud, rather than which intercept you want, prevents the swap almost entirely.
Figure (svg): A line crossing both axes with the two intercepts marked, showing that the x-intercept is found by setting the output to zero and the y-intercept by setting the input to zero
OpenStax, Precalculus, §2.2 Graphs of Linear Functions §2.2, pp. 212-218
Picture it
Each axis is defined by one coordinate being zero, which is what makes the method work.
Figure (svg): A line crossing both axes with the two intercepts marked, showing that the x-intercept is found by setting the output to zero and the y-intercept by setting the input to zero
The x-intercept required solving an equation and the y-intercept only an evaluation. That asymmetry is why the y-intercept is the one that appears for free in slope-intercept form.
Worked example
Two substitutions, each followed by a one-step solve.
\[ \text{Find both intercepts of } 3x+4y=12. \]
Set x to zero
Why: This finds where it meets the vertical axis.
\[ 4 y = 12 \]
Solve
Why: Divide by 4.
\[ y = 3,\text{ point } (0, 3) \]
Set y to zero
Why: This finds where it meets the horizontal axis.
\[ 3 x = 12 \]
Solve
Why: Divide by 3.
\[ x = 4,\text{ point } (4, 0) \]
Figure (svg): A line crossing both axes with the two intercepts marked, showing that the x-intercept is found by setting the output to zero and the y-intercept by setting the input to zero
\[ (0,\,3) \quad \text{and} \quad (4,\,0) \]
Verify: check both satisfy the original
Why: At (0, 3): zero plus 12 is 12. At (4, 0): 12 plus zero is 12. Both work. This is the fastest way to graph a line given in standard form, since neither intercept required rearranging the equation at all.
OpenStax, Precalculus, §2.2 Graphs of Linear Functions §2.2, pp. 213-215
Discrimination
The pairing is the opposite of what the names suggest.
Sort into buckets
Sort each action by what it produces.
Worked example
Horizontal lines are the exception, and the algebra says so.
\[ \text{Find the intercepts of } y = 4. \]
Set x to zero
Why: The rule ignores the input.
\[ y = 4 \]
Read the y-intercept
Why: It meets the vertical axis at 4.
\[ \text{point } (0, 4) \]
Set y to zero
Why: The equation becomes a false statement.
\[ 4 = 0 \]
Interpret
Why: No input satisfies it.
Figure (svg): The solution to Worked example a line with no x-intercept shown as a ladder of expressions, one row per legal move
\[ (0,4); \text{ no } x\text{-intercept} \]
Verify: confirm from the picture
Why: A horizontal line at height 4 runs parallel to the horizontal axis and never touches it, so there is genuinely nowhere to cross. The equation four equals zero is the algebra reporting that impossibility, and a contradiction like that is always worth reading as 'no solutions' rather than as a mistake.
OpenStax, Precalculus, §2.2 Graphs of Linear Functions §2.2, pp. 216-217
Error analysis
A student finds the x-intercept of a line.
Annotate
On: \( y=2x-6: \quad x\text{-intercept} \;\Longrightarrow\; \text{set } x=0 \;\Longrightarrow\; y=-6 \)
Name the substitution rather than the target: say 'I am setting y to zero', and the intercept you get is the one you wanted.
Prediction
A line has slope zero and passes through the point at height 5.
Predict first
How many x-intercepts does it have?
Correct: None, since it never reaches height zero.
Why: A horizontal line at height 5 stays at height 5 forever, so it never crosses the horizontal axis. The infinitely-many answer would apply to the horizontal axis itself, which is the one horizontal line that lies along the axis and meets it everywhere.
Faded example
Find where the line with rule 5x minus 2y equal to 20 crosses the horizontal axis.
Fill in the blanks
\text20 y = 0: \quad 5x = 4 \;\Longrightarrow\; x = ___
Why: Setting y to zero removes the middle term, leaving 5x equal to 20 and so x equal to 4. The intercept is the point with coordinates 4 and 0. Checking: 20 minus 0 is 20, as required.
Socratic
Slope-intercept form displays the y-intercept and says nothing about the x-intercept.
Discussion prompt
Explain why the two intercepts are not equally easy to read off.
Hint: Which one requires solving an equation?
Answer:
The y-intercept is an evaluation: substitute zero for the input and compute. Every term with an x in it vanishes, so what is left is the constant, which is already written down.
The x-intercept requires solving: the output is known and the input is not, which is §1.1's distinction between evaluating and solving. Solving is harder than evaluating, and no rearrangement of the formula can make it free.
That asymmetry runs through the whole course. Reading off where a graph starts is always easy; finding where it crosses zero is the hard question, and for polynomials in Chapter 3 it becomes the central problem of the chapter.
Section
Section 3
Concept
Two distinct lines are parallel exactly when they have the same slope. Since slope is the line's direction, equal slopes mean the lines never converge.
The vertical case is a genuine exception rather than a technicality: two vertical lines are obviously parallel and the slope criterion cannot say so, because neither slope exists. It is one more reason vertical lines have to be handled separately throughout this chapter.
Figure (svg): Two lines with the same slope and different intercepts drawn on the same axes, never meeting, with their slope triangles shown identical
OpenStax, Precalculus, §2.2 Graphs of Linear Functions §2.2, pp. 218-223
Picture it
The dashed triangles on the two lines have the same rise and the same run.
Figure (svg): Two lines with the same slope and different intercepts drawn on the same axes, never meeting, with their slope triangles shown identical
The lines maintain a constant separation because they advance at the same rate. Any difference in slope, however small, would eventually bring them together.
Worked example
Copy the slope, then use the given point.
\[ \text{Find the line through } (2,7) \text{ parallel to } y=3x-4. \]
Read the slope of the given line
Why: It is already in slope-intercept form.
\[ m = 3 \]
Copy it
Why: Parallel means the same slope.
\[ \text{new slope is } 3 \]
Use point-slope with the given point
Why: Substitute directly.
\[ y - 7 = 3(x - 2) \]
Expand and solve for y
Why: Distribute and collect.
\[ y = 3 x + 1 \]
Figure (svg): The solution to Worked example write a parallel line shown as a ladder of expressions, one row per legal move
\[ y = 3x+1 \]
Verify: check the point and the parallelism
Why: At the input 2 the answer gives 6 plus 1, which is 7 — the given point. The slopes are both 3 and the intercepts differ, 1 against negative 4, so the lines are genuinely parallel and genuinely distinct. Both checks are needed: matching intercepts would have meant the same line.
OpenStax, Precalculus, §2.2 Graphs of Linear Functions §2.2, pp. 219-221
Sorting
Compare both the slopes and the intercepts.
Sort into buckets
Sort each pair of lines.
Worked example
Rearrange both before comparing anything.
\[ \text{Are } 6x-2y=8 \text{ and } y=3x+1 \text{ parallel?} \]
Rearrange the first
Why: Isolate y.
\[ -2 y = -6 x + 8 \]
Divide through
Why: Both terms by negative 2.
\[ y = 3 x - 4 \]
Compare the slopes
Why: Both are 3.
Compare the intercepts
Why: Negative 4 against 1, which differ.
Figure (svg): The solution to Worked example are these two lines parallel shown as a ladder of expressions, one row per legal move
\[ \text{Parallel: both have slope } 3 \text{ and different intercepts.} \]
Verify: check they really never meet
Why: Setting the two rules equal gives 3x minus 4 equal to 3x plus 1, and cancelling the 3x leaves negative 4 equal to 1, which is false. No input satisfies both, so there is no crossing point — which is what parallel means, arrived at algebraically.
OpenStax, Precalculus, §2.2 Graphs of Linear Functions §2.2, pp. 221-222
Trap
\[ y=2x+5 \quad \text{and} \quad 4x-2y=-10 \;\Longrightarrow\; \text{parallel} \]
Rearrange the second and compare slopes
Why: Both come out with slope 2, so the lines are declared parallel.
The intercepts are not compared, and the pair is reported as two parallel lines.
They are the same line. Rearranging the second gives y equals 2x plus 5, which is the first equation exactly.
Parallel requires the lines to be distinct. Two copies of one line meet at every point, which is the opposite of never meeting.
Compare the intercepts as well as the slopes. Equal slopes and equal intercepts is the coincident case, and its system has infinitely many solutions rather than none — an important distinction in Chapter 9.
Prediction
Two distinct parallel lines are written as a system of equations.
Predict first
How many solutions does the system have?
Correct: None, since the lines never meet.
Why: A solution is a point on both lines, and distinct parallel lines share no point. Algebraically, setting the rules equal cancels the x terms and leaves a false statement about the constants. Infinitely many would be the coincident case, where the two equations describe the same line.
Faded example
Find the line through the point at (negative 1, 4) parallel to the line with slope negative 2.
Fill in the blanks
y - 4 = -2(x + 1) \;\Longrightarrow\; y = -2x + 2
Why: The slope is copied unchanged, and the point supplies the two subscripted values. The input coordinate negative 1 becomes a plus 1 inside the bracket. Expanding gives negative 2x minus 2 plus 4, which is negative 2x plus 2, and checking at the input negative 1 gives 4 as required.
Edge cases
The slope criterion for parallelism cannot handle vertical lines.
Discussion prompt
Are two vertical lines parallel, and what does that say about the criterion?
Hint: Do they ever meet?
Answer:
Yes, they are parallel: two distinct vertical lines never meet, which is what parallel means geometrically.
But the slope criterion cannot say so, because neither line has a slope. Equal slopes is a sufficient and necessary condition only among lines that have slopes at all.
This is why careful statements say 'two non-vertical lines are parallel exactly when their slopes are equal', and treat the vertical case separately. The geometric definition — never meeting — is the more general one, and the slope criterion is a convenient test that happens to cover every case but one.
Section
Section 4
Concept
Two lines with slopes are perpendicular exactly when the product of their slopes is negative one — equivalently, when each slope is the negative reciprocal of the other.
\[ m_1 m_2 = -1 \;\Longleftrightarrow\; m_2 = -\frac{1}{m_1} \]
The geometric reason is worth carrying. Rotating a slope triangle by a quarter turn turns its run into a rise and its rise into a run, and reverses the direction of one of them. Swapping rise and run inverts the fraction; reversing one direction changes the sign. That is the negative reciprocal, derived rather than memorised.
Figure (svg): Two perpendicular lines with their slope triangles drawn, showing that rotating one triangle by a quarter turn swaps its rise and run and reverses one sign, which is why the slopes are negative reciprocals
OpenStax, Precalculus, §2.2 Graphs of Linear Functions §2.2, pp. 223-229
Picture it
The two triangles are the same triangle in two orientations.
Figure (svg): Two perpendicular lines with their slope triangles drawn, showing that rotating one triangle by a quarter turn swaps its rise and run and reverses one sign, which is why the slopes are negative reciprocals
Up 2 and right 1 becomes right 2 and down 1. Rise and run have traded places and one has changed sign, which is exactly what the negative reciprocal does.
Worked example
Flip and negate, then use the point.
\[ \text{Find the line through } (4,1) \text{ perpendicular to } y=\tfrac{2}{3}x+5. \]
Read the given slope
Why: It is two thirds.
\[ m = \frac{2}{3} \]
Flip the fraction
Why: Reciprocal of two thirds.
\[ \frac{3}{2} \]
Change the sign
Why: The perpendicular slope.
\[ -\frac{3}{2} \]
Use point-slope with the given point
Why: Substitute and expand.
\[ y - 1 = -(\frac{3}{2}) (x - 4) \]
Figure (svg): Two perpendicular lines with their slope triangles drawn, showing that rotating one triangle by a quarter turn swaps its rise and run and reverses one sign, which is why the slopes are negative reciprocals
\[ y = -\tfrac{3}{2}x + 7 \]
Verify: multiply the slopes
Why: Two thirds times negative three halves gives negative one, which is the condition. And at the input 4 the answer gives negative 6 plus 7, which is 1 — the given point. Both the perpendicularity and the point check out, which is everything the question asked for.
OpenStax, Precalculus, §2.2 Graphs of Linear Functions §2.2, pp. 225-227
Matching
Flip the fraction and change the sign.
Match the pairs
Why: In each case the fraction is inverted and the sign reversed. The second shows a negative slope pairing with a positive one, and the fourth is the self-similar case where the partner of negative one is positive one — the two diagonals, which are indeed perpendicular. Every pair multiplies to negative one, which is the check.
Worked example
Rearrange both, then multiply.
\[ \text{Are } 2x+5y=10 \text{ and } y=\tfrac{5}{2}x-1 \text{ perpendicular?} \]
Rearrange the first
Why: Isolate y.
\[ 5 y = -2 x + 10 \]
Divide through
Why: Both terms by 5.
\[ y = -(\frac{2}{5}) x + 2 \]
Read both slopes
Why: Negative two fifths and five halves.
\[ -\frac{2}{5}\text{ and } \frac{5}{2} \]
Multiply them
Why: The product decides it.
\[ -\frac{10}{10} = -1 \]
Figure (svg): The solution to Worked example test two lines for perpendicularity shown as a ladder of expressions, one row per legal move
\[ -\tfrac{2}{5}\cdot\tfrac{5}{2} = -1: \text{ perpendicular} \]
Verify: notice which operations were performed
Why: Going from negative two fifths to five halves, the fraction was inverted and the sign flipped — both operations. Doing only the inversion would give negative five halves, and only the sign flip would give two fifths, and neither of those is perpendicular to the original. Checking the product is the reliable test because it catches a half-done conversion.
OpenStax, Precalculus, §2.2 Graphs of Linear Functions §2.2, pp. 227-229
Error analysis
A student finds the slope perpendicular to three quarters.
Annotate
On: \( m = \tfrac{3}{4} \;\Longrightarrow\; m_\perp = -\tfrac{3}{4} \)
Both operations, every time: flip the fraction and change the sign. Then multiply the two slopes together as a check — it is one line and it is decisive.
Prediction
A line is horizontal, with slope zero.
Predict first
What is the slope of a line perpendicular to it?
Correct: Undefined, since that line is vertical.
Why: A line perpendicular to a horizontal one is vertical, and a vertical line has no slope. The product rule cannot handle this pair, since it would require multiplying zero by an undefined quantity. It is the same exception as with parallel vertical lines: the geometry is clear and the slope criterion runs out.
Faded example
A line has slope negative five sixths. Find the slope perpendicular to it.
Fill in the blanks
\text6/5 -\tfrac-1___ \;\Longrightarrow\; \text___ ___, \qquad \text___ -\tfrac______\cdot___ = ___
Why: Inverting gives negative six fifths, and negating that gives positive six fifths. The check multiplies negative five sixths by six fifths, and the fives and sixes cancel to leave negative one. Doing the check every time is what distinguishes a reliably correct answer from one that is right half the time.
Explain it to yourself
The condition looks arbitrary until the picture is drawn.
Discussion prompt
Explain why perpendicular slopes are negative reciprocals, using a slope triangle.
Hint: What does rotating the triangle by a quarter turn do to its rise and run?
Answer:
Draw the slope triangle for a line: over by the run, up by the rise. Now rotate the whole triangle a quarter turn, which is what making a line perpendicular does to its direction.
After the rotation, what was the run points vertically and what was the rise points horizontally: rise and run have traded places, which inverts the fraction. And one of the two now points the opposite way, which reverses the sign.
Together those give the negative reciprocal. So the rule is not a convention to memorise but a description of what a quarter turn does — which is also why it is the one relationship in this section that cannot be guessed from the algebra alone.
Section
Section 5
Concept
A point on both lines has the same input and the same output in each. Setting the two rules equal finds the input where that happens, and substituting back gives the output.
This is a system of two equations in two unknowns, solved here by the substitution method without that name being used. Chapter 9 develops the machinery for larger systems, but the two-line case is complete already, and the three possible outcomes there are exactly the three seen here.
Figure (svg): Two lines crossing at a single point, with the point marked and the two equations shown, illustrating that solving them simultaneously finds where they meet
OpenStax, Precalculus, §2.2 Graphs of Linear Functions §2.2, pp. 229-232
Picture it
The point marked lies on both lines, so it satisfies both equations.
Figure (svg): Two lines crossing at a single point, with the point marked and the two equations shown, illustrating that solving them simultaneously finds where they meet
Two lines with different slopes converge and cross exactly once. Making the slopes equal removes the crossing entirely, which is the geometric meaning of a system with no solution.
Worked example
Equate, solve, substitute back.
\[ \text{Where do } y=2x-3 \text{ and } y=-x+6 \text{ cross?} \]
Set the two rules equal
Why: At the crossing the outputs agree.
\[ 2 x - 3 = -x + 6 \]
Collect the x terms
Why: Add x to both sides.
\[ 3 x - 3 = 6 \]
Solve for the input
Why: Add 3 and divide by 3.
\[ x = 3 \]
Substitute back
Why: Into either rule.
\[ y = 2(3) - 3 = 3 \]
Figure (svg): Two lines crossing at a single point, with the point marked and the two equations shown, illustrating that solving them simultaneously finds where they meet
\[ (3,\,3) \]
Verify: check in the OTHER equation too
Why: The second rule at the input 3 gives negative 3 plus 6, which is also 3. Both rules agree there, so the point really is on both lines. Substituting into only the rule you already used would confirm nothing, since that equation was used to find the answer.
OpenStax, Precalculus, §2.2 Graphs of Linear Functions §2.2, pp. 230-231
Sorting
Compare the slopes first, then the intercepts if needed.
Sort into buckets
Sort each pair.
Worked example
A contradiction is an answer, not a failure.
\[ \text{Where do } y=4x+1 \text{ and } y=4x-7 \text{ cross?} \]
Set them equal
Why: As usual.
\[ 4 x + 1 = 4 x - 7 \]
Collect the x terms
Why: Subtract 4x from both sides.
\[ 1 = -7 \]
Read the result
Why: A statement with no variable in it, and false.
Interpret
Why: No input can make it true.
Figure (svg): The solution to Worked example when the algebra says no shown as a ladder of expressions, one row per legal move
\[ \text{No solution: the lines are parallel.} \]
Verify: confirm from the slopes
Why: Both slopes are 4 and the intercepts differ, so the lines are parallel and distinct — which is exactly the no-crossing case. Had the intercepts also matched, the algebra would have ended in a true statement such as 1 equals 1, meaning every point works and the lines coincide.
OpenStax, Precalculus, §2.2 Graphs of Linear Functions §2.2, pp. 231-232
Trap
\[ 2x-3 = -x+6 \;\Longrightarrow\; x=3 \;\Longrightarrow\; \text{the crossing point is } 3 \]
Solve for x and report it
Why: The algebra is carried out correctly and 3 is the right input.
The answer is given as the single number 3.
A crossing point has two coordinates. The input is 3, and the output must be found by substituting back, giving 3 as well — so the point is at (3, 3).
Reporting only the input answers half the question, and the two coordinates are usually different numbers, so the omission is normally visible.
Substitute back and state a point. Then check that point in the equation you did NOT use to find it, which is the step that catches arithmetic slips.
Prediction
Two equations describing the same line are set equal to each other.
Predict first
What does the algebra produce?
Correct: A true statement with no variable, such as 5 = 5.
Why: Both the variable terms and the constants cancel, leaving something that is true regardless of x — which means every input works and the lines coincide. A false statement would mean no input works, giving parallel lines. Distinguishing these two outcomes is the whole skill, since both look like the variable has vanished.
Faded example
Find where the lines with rules 5x plus 2 and 3x plus 8 cross.
Fill in the blanks
5x + 2 = 3x + 8 \;\Longrightarrow\; 2x = 6 \;\Longrightarrow\; x = 3, \; y = 17
Why: Subtracting 3x and 2 from both sides gives 2x equal to 6, so x is 3. Substituting into the first rule gives 15 plus 2, which is 17, and the second gives 9 plus 8, which is also 17 — so the point at (3, 17) is on both lines, confirmed in both equations.
Real world
A break-even calculation is exactly this computation.
Discussion prompt
A business has costs of 500 plus 3 per unit and revenue of 8 per unit. What does the crossing point mean, and what do the regions on either side mean?
Hint: Which line is higher on each side of the crossing?
Answer:
Setting 500 plus 3 times the units equal to 8 times the units gives 5 times the units equal to 500, so the crossing is at 100 units. That is the break-even point: costs and revenue are equal there.
To the left of it, fewer than 100 units, the cost line is above the revenue line, so the business loses money. To the right, revenue is above cost and it makes a profit.
The slopes carry the meaning too. Revenue rises at 8 per unit and cost at only 3, so revenue gains 5 per unit on cost — which is why the lines converge and cross despite cost starting 500 higher. Break-even analysis is this section applied without renaming anything.
Comparison
Fill the blanks from memory. Two relationships, two conditions, and only one of them is obvious.
Comparison matrix
| parallel | perpendicular | |
|---|---|---|
| slope condition | the slopes are equal | the product of the slopes is -1 |
| how to build one | copy the slope | flip the fraction and change the sign |
| how many crossings | none, unless they coincide | exactly one |
| the special pair | two vertical lines | one horizontal and one vertical |
| why the criterion fails there | neither has a slope | one slope is undefined, so no product exists |
Both special pairs involve a vertical line, which is the recurring exception of this chapter: geometrically unremarkable, algebraically outside every formula.
Pattern
Whether the relationship is parallel or perpendicular, the route is the same.
Step 3 costs one multiplication and catches the commonest error in the section, which is performing only one of the two operations that make a negative reciprocal.
OpenStax Algebra and Trigonometry 2e, §4.1 Linear Functions §4.1
Check
Set the other variable to zero.
Check your understanding
What is the x-intercept of the line 4x - 5y = 20?
Answer: A
Why: Setting y to zero gives 4x equal to 20, so x is 5 and the point is at (5, 0). Checking: 20 minus 0 is 20, as required.
Check
Flip and negate.
Check your understanding
What slope is perpendicular to a line of slope -4/7?
Answer: A
Why: Inverting gives negative seven quarters, and negating gives positive seven quarters. The check: negative four sevenths times seven quarters is negative one.
Check
Read the outcome of the algebra.
Check your understanding
Setting two linear rules equal produces the statement 6 = 6. What does that mean?
Answer: A
Why: Both the variable terms and the constants cancelled, leaving a statement that is true regardless of the input. Every input satisfies both rules, so the two equations describe the same line and there are infinitely many solutions.
Real world
The perpendicular condition is what every drawing program uses to snap a line square to another.
Discussion prompt
A drafting program lets a user draw a line perpendicular to an existing one. What must it compute, and what case must it special-case in the code?
Hint: Which slope has no negative reciprocal?
Answer:
It computes the negative reciprocal of the existing line's slope and draws through the chosen point — precisely this section's procedure, executed thousands of times a second.
The case that must be special-cased is the horizontal line, whose slope is zero. Its negative reciprocal would require dividing by zero, so the code cannot compute it and must instead recognise that the perpendicular is vertical.
The vertical line has to be special-cased for the same reason in the other direction, since it has no slope to invert. This is the everyday consequence of the exception this chapter keeps meeting: vertical lines are geometrically ordinary and algebraically outside the formulas, so any implementation has to handle them by hand.
Commit first
State your confidence along with your answer.
Predict first
Two lines have slopes 3 and negative 3. What is their relationship?
Correct: They cross, but are not perpendicular.
Why: The slopes differ, so the lines cross exactly once. But the product is negative 9, not negative 1, so they are not perpendicular — negating a slope is only half of what perpendicularity requires, and the reciprocal step is missing. Slopes of 3 and negative one third would be perpendicular.
Explain it
The test of understanding is being able to say why, not just what.
Discussion prompt
Explain to a classmate why perpendicular slopes are negative reciprocals rather than just negatives, using a picture rather than the formula.
Hint: Draw a slope triangle and turn the page.
Answer:
Draw the slope triangle for a line — say up 2 and right 1. Now turn the page a quarter turn, which is what making a line perpendicular does.
The triangle is now 2 across and 1 up-or-down: the rise and the run have swapped, which turns 2 over 1 into 1 over 2. And the vertical leg now points the other way, which is where the minus sign comes from.
Explaining why 'just negating' fails is the useful half. Negating alone keeps the triangle's shape and only mirrors it, which produces a line at the wrong angle — the two would form a symmetric V rather than a right angle. Both operations, because the rotation does both.
Exit ticket
One honest answer, so the next lesson can start in the right place.
Predict first
Which idea from this lesson would you most want to see again?
Correct: Any of these is a legitimate answer; the useful one is the honest one.
Why: There is no correct choice here. The fourth is the one that recurs furthest into the course, appearing again for tangent and normal lines and throughout Chapter 10, so it repays attention now. The second costs marks for reasons that are purely about keeping two similar procedures apart.
Connect it up
One page, drawn from memory, is worth more than rereading the section.
Draw it
Draw one line, then draw a second parallel to it and a third perpendicular to it, all on the same axes. Label all three slopes and show the arithmetic that produces the second and third from the first. Then mark the crossing point of the first and third, and write the equation you would solve to find it.
If the two perpendicular slopes multiply to negative one on your page, and the parallel pair have different intercepts, you have both relationships and the trap in each.
Recap
Five things, and the fourth is the one that recurs longest.
| if you remember one thing | it should be this |
|---|---|
| about graphing | plot the intercept and walk the slope; plot a third point to check |
| about intercepts | set the OTHER variable to zero |
| about parallel | equal slopes, and check the intercepts differ |
| about perpendicular | flip AND negate, then multiply to confirm negative one |
Section 2.3 puts these lines to work on real quantities, where the slope has units and the domain is limited by what the situation allows.
OpenStax, Precalculus, §2.2 Graphs of Linear Functions §2.2, pp. 205-232 — everything on these slides traces back here
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