Writing a linear equation from slope and intercept, from slope and a point using point-slope form, and from two points; finding lines parallel or perpendicular to a given line through a given point; and building a linear model from two data values.
Subject: Algebra 2 · 65 slides · symbolic lesson
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Title
Algebra 2 · Chapter 2 — Linear Equations and Functions
Write Equations of Lines
Objectives
Five outcomes. The second is the tool that makes the other three possible.
McDougal Littell Algebra 2 (Texas Edition), Ch. 2 Linear Equations and Functions — Lesson 2.4 Write Equations of Lines §2.4, pp. 98-103 — the lesson these objectives are drawn from
Warm-up
Lesson 2.3 read m and b off an equation and drew the line. This lesson runs the same machine backwards.
Discussion prompt
A line crosses the vertical axis at negative two and rises three units for every four across. Write its equation without any formula — just by naming the two numbers slope-intercept form asks for.
Hint: You need m and b, and you have both.
Answer:
\[ m = \tfrac{3}{4}, \quad b = -2 \;\Longrightarrow\; y = \tfrac{3}{4}x - 2 \]
That is the whole of the first case. The rest of this lesson exists because you are usually not handed the intercept — you are handed a point that is not on the vertical axis, or two points, and the equation has to be built from those instead.
Concept
Writing the equation of a line always needs the same two facts: how steep it is, and where it is. What changes is how those facts arrive. Slope with an intercept goes straight into slope-intercept form; slope with any other point needs point-slope form; two points need the slope computed first.
point-slope form — The form y minus y sub one equals m times the quantity x minus x sub one, where m is the slope and the point with coordinates x sub one and y sub one lies on the line.
In this book every answer is simplified to slope-intercept form at the end, so point-slope is a working form rather than a final one.
Figure (svg): Three boxes showing what you are given and which form to use: slope and intercept, slope and a point, or two points
McDougal Littell Algebra 2 (Texas Edition), Ch. 2 Linear Equations and Functions — Lesson 2.4 Write Equations of Lines §2.4, pp. 98-98 — Writing an Equation of a Line
Section
Section 1
Concept
When you know the slope and the y-intercept, the equation is a substitution into y equals m x plus b and nothing more. Reading those two numbers off a graph is the only skill involved.
\[ y = mx + b \]
Read the intercept where the line crosses the vertical axis, and read the slope by stepping from one lattice point to the next: count the rise, then the run.
Figure (svg): Three boxes showing what you are given and which form to use: slope and intercept, slope and a point, or two points
McDougal Littell Algebra 2 (Texas Edition), Ch. 2 Linear Equations and Functions — Lesson 2.4 Write Equations of Lines §2.4, pp. 98-98
Picture it
The whole lesson on one card.
Figure (svg): Three boxes showing what you are given and which form to use: slope and intercept, slope and a point, or two points
Notice the third row does not introduce a new form. Two points give you a slope, and once you have a slope you are back in the second row.
Worked example
Example 1. The graph gives both numbers directly.
\[ \text{A line crosses the vertical axis at } -2 \text{ and rises } 3 \text{ for every } 4 \text{ across. Write its equation.} \]
Read the y-intercept off the graph
Why: The line crosses the vertical axis at negative two, so b is negative two.
\[ b = -2 \]
Read the slope by counting rise over run
Why: From the crossing point, going right four lands three higher.
\[ m = \frac{3}{4} \]
Substitute both into slope-intercept form
Why: There is nothing to solve; the form is already arranged for these two numbers.
\[ y = (\frac{3}{4}) x + (-2) \]
Simplify the sign
Why: Plus negative two is minus two.
\[ y = (\frac{3}{4}) x - 2 \]
Figure (svg): The solution to Worked example read the equation off a graph shown as a ladder of expressions, one row per algebraic move
\[ y = \tfrac{3}{4}x - 2 \]
Verify: test a second lattice point from the graph
Why: Stepping right four and up three from (0,-2) gives (4,1). Substituting: three quarters of four is three, minus two is one — correct. Checking a point other than the intercept is what confirms the slope as well as the intercept.
McDougal Littell Algebra 2 (Texas Edition), Ch. 2 Linear Equations and Functions — Lesson 2.4 Write Equations of Lines §2.4, pp. 98-98
Matching
Four pairs, four equations.
Match the pairs
Why: The pairs are deliberately arranged as two swapped couples, so the only way through is to attach each number to its role rather than to its position. The slope always multiplies x; the intercept always stands alone. Substituting x equal to zero recovers b in every case, which is the fastest way to check you did not swap them.
Worked example
Guided Practice 1 through 3. Pure substitution, including a fractional pair.
\[ \text{Write the equation given: (a) } m = 3, b = 1; \; \text{(b) } m = -2, b = -4; \; \text{(c) } m = -\tfrac{3}{4}, b = \tfrac{7}{2}. \]
Substitute the first pair
Why: Three for m, one for b.
\[ y = 3 x + 1 \]
Substitute the second pair
Why: Negative two for m, negative four for b, and plus negative four is minus four.
\[ y = -2 x - 4 \]
Substitute the third pair
Why: Negative three quarters for m, seven halves for b.
\[ y = -(\frac{3}{4}) x + \frac{7}{2} \]
Notice that fractions need no special treatment
Why: The form does not care whether m and b are integers; only the substitution matters.
Figure (svg): The solution to Worked example three from the guided practice shown as a ladder of expressions, one row per algebraic move
\[ y = 3x + 1; \quad y = -2x - 4; \quad y = -\tfrac{3}{4}x + \tfrac{7}{2} \]
Verify: evaluate each at x equal to zero
Why: All three give back their own b: 1, negative 4, and seven halves. Since x equal to zero is the vertical axis, that is exactly what the y-intercept means, so the substitution went in the right slots.
McDougal Littell Algebra 2 (Texas Edition), Ch. 2 Linear Equations and Functions — Lesson 2.4 Write Equations of Lines §2.4, pp. 98-98
Trap
\[ m = -2, \quad b = -4 \]
Write the two numbers in the order they were given
Why: The form is filled in left to right rather than by what each letter means.
\[ y = -4x - 2 \quad \text{(wrong)} \]
This line crosses the vertical axis at negative two, but the intercept was supposed to be negative four.
\[ m = -2, \quad b = -4 \]
Put the SLOPE next to the x and the INTERCEPT on its own
Why: In y equals m x plus b, m multiplies x and b stands alone.
\[ y = -2x - 4 \]
Check at x equal to zero: the equation gives negative four, which is the intercept as required. That one substitution catches the swap every time.
Reverse engineer
Read the two numbers back out.
Fill in the blanks
y = -\tfrac-\frac{3}{4}___x + \tfrac______ \;\Longrightarrow\; m = ___, \; b = \tfrac______
Why: The slope is the coefficient of x with its sign, so it is negative three quarters — the line falls three for every four across. The intercept of seven halves means it crosses the vertical axis at 3.5. Reading in this direction is Lesson 2.3's skill; reading in the other direction is this lesson's, and they are the same fact used two ways.
Prediction
Commit before reasoning.
Predict first
Two lines have the same slope but different y-intercepts. What is true of them?
Correct: They are parallel.
\[ y = 3x + 1 \quad \text{and} \quad y = 3x - 4 \]
Why: Equal slopes with different intercepts is exactly Lesson 2.2's definition of parallel: the lines are equally steep and therefore never meet, but they are genuinely different lines because they cross the vertical axis in different places. Equal slopes AND equal intercepts would make them the same line, which is why the definition of parallel requires the lines to be distinct.
Explain it to yourself
One case needs no formula at all.
\[ y = mx + b \]
Discussion prompt
Explain why being given the slope and the y-intercept is the easiest of the three cases, in terms of what the form y equals m x plus b already contains. What is missing in the other two cases that this one hands you for free?
Hint: Ask what the form is arranged to accept.
Answer:
Slope-intercept form has one slot for the slope and one for the intercept, so being given exactly those two numbers means the form can be filled in with no algebra.
In the other cases the intercept is not given. You are handed a point that is somewhere else on the line, and the intercept has to be worked out — which is precisely what point-slope form does for you without your having to find it explicitly.
Section
Section 2
Concept
If a line has slope m and passes through a known point, then for any other point on it the slope formula must hold. Multiplying that formula through by the denominator gives point-slope form, with nothing in a fraction.
\[ y - y_1 = m(x - x_1) \]
Nothing new is being asserted. The form is convenient because it accepts a slope and any point at all, whereas slope-intercept form only accepts the one point on the vertical axis.
Figure (svg): The point-slope form derived from the slope formula, with a line through a known point and a general point
McDougal Littell Algebra 2 (Texas Edition), Ch. 2 Linear Equations and Functions — Lesson 2.4 Write Equations of Lines §2.4, pp. 98-99 — point-slope form
Picture it
Take the slope formula with one point known and one point general.
Figure (svg): The point-slope form derived from the slope formula, with a line through a known point and a general point
The subscripts mark the point you were given; the plain x and y stand for any point on the line. Getting those two roles the right way round is the only thing to be careful about.
Worked example
Example 2. Substitute, distribute, simplify.
\[ \text{Write the equation of the line through } (5, 4) \text{ with slope } -3. \]
Choose point-slope form
Why: You have a slope and a point that is not the intercept, which is exactly what this form accepts.
\[ y - y 1 = m(x - x 1) \]
Substitute the slope and the point
Why: Negative three for m, five for x sub one, four for y sub one.
\[ y - 4 = -3(x - 5) \]
Distribute
Why: Negative three times negative five is positive fifteen.
\[ y - 4 = -3 x + 15 \]
Add 4 to both sides to reach slope-intercept form
Why: Fifteen plus four is nineteen.
\[ y = -3 x + 19 \]
Figure (svg): The solution to Worked example slope and a point shown as a ladder of expressions, one row per algebraic move
\[ y = -3x + 19 \]
Verify: substitute the given point into the final equation
Why: At x equal to 5: negative three times five is negative fifteen, plus nineteen is four — which is the y coordinate given. The point really does lie on the line, and the slope of negative three is visible as the coefficient.
McDougal Littell Algebra 2 (Texas Edition), Ch. 2 Linear Equations and Functions — Lesson 2.4 Write Equations of Lines §2.4, pp. 99-99
Fill the middle
A line through (-3, 5) with slope negative two.
Fill in the blanks
y - 5 = -2\left(x + 3\right) \;\Longrightarrow\; y = -2x - 1
Why: Minus negative three is plus three, so the bracket is x plus three. Distributing gives negative 2x minus 6, and adding 5 gives y equals negative 2x minus 1. Checking the point: at x equal to negative three, negative two times negative three is six, minus one is five — correct.
Worked example
Guided Practice 4. The double negative is where this goes wrong.
\[ \text{Write the equation of the line through } (-1, 6) \text{ with slope } 4. \]
Substitute into point-slope form
Why: Four for m, negative one for x sub one, six for y sub one. Keep the brackets.
\[ y - 6 = 4(x - (-1)) \]
Simplify the double negative inside the bracket
Why: Minus negative one is plus one, so the bracket becomes x plus one.
\[ y - 6 = 4(x + 1) \]
Distribute
Why: Four times x is 4x; four times one is four.
\[ y - 6 = 4 x + 4 \]
Add 6 to both sides
Why: Four plus six is ten.
\[ y = 4 x + 10 \]
Figure (svg): The solution to Worked example a negative coordinate shown as a ladder of expressions, one row per algebraic move
\[ y = 4x + 10 \]
Verify: substitute the given point
Why: At x equal to negative one: four times negative one is negative four, plus ten is six — the given y coordinate. Note how the sign flipped inside the bracket while the coordinate itself stayed negative; writing the substitution with brackets is what keeps those two straight.
McDougal Littell Algebra 2 (Texas Edition), Ch. 2 Linear Equations and Functions — Lesson 2.4 Write Equations of Lines §2.4, pp. 99-99
Trap
\[ \text{Through } (-1, 6), \; m = 4. \]
Write the point's coordinates straight into the form
Why: The minus signs in the form are copied over as if they were part of the coordinates.
\[ y - 6 = 4(x - 1) \quad \text{(wrong)} \]
\[ y = 4x + 2 \]
Substituting x equal to negative one gives negative two, not six. The point is not on this line.
\[ \text{Through } (-1, 6), \; m = 4. \]
Substitute with brackets, then simplify the signs
Why: The form contains x minus x sub one, and x sub one is negative one, so the bracket is x minus negative one.
\[ y - 6 = 4(x - (-1)) = 4(x + 1) \]
\[ y = 4x + 10 \]
The check is one substitution: the given point must satisfy the final equation, and it does.
Sorting
Do not write any equations. Just choose the form.
Sort into buckets
Sort each set of given information by the form it calls for.
The fifth item is worth noticing: a point with x equal to zero IS the y-intercept, so it belongs in the first bucket even though it was written as a point.
Error analysis
A student writes the equation through (4, -2) with slope 3.
Annotate
On: \( \begin{aligned} y - (-2) &= 3(x - 4) \\ y + 2 &= 3x - 12 \\ y &= 3x - 10 \end{aligned} \)
Always finish by substituting the point you were given. It is the one check that tests the whole chain rather than any single step.
Explain it
A classmate has memorised point-slope form and thinks it is arbitrary.
Discussion prompt
In three sentences, derive it for them from the slope formula they already know, and say why it is more useful than solving for b every time.
Hint: Start from m equals the difference quotient with one point general.
Answer:
\[ m = \frac{y - y_1}{x - x_1} \;\Longrightarrow\; m(x - x_1) = y - y_1 \]
It is the slope formula with the denominator cleared, so it says nothing new — only that the slope between the known point and any other point on the line is m. It is more useful than solving for b because it accepts any point at all, so you never have to find the intercept as a separate step.
Section
Section 3
Concept
These problems hand you a line and a point. The line supplies the slope — equal for parallel, negative reciprocal for perpendicular — and the point supplies the position. Then it is an ordinary point-slope problem.
\[ \text{parallel: } m_2 = m_1 \qquad \text{perpendicular: } m_2 = -\tfrac{1}{m_1} \]
The given line's own intercept is irrelevant. Only its slope is used, because the new line's position comes entirely from the given point.
Figure (svg): One point with two lines through it, one parallel to a given line and one perpendicular to it
McDougal Littell Algebra 2 (Texas Edition), Ch. 2 Linear Equations and Functions — Lesson 2.4 Write Equations of Lines §2.4, pp. 99-99 — Write equations of parallel or perpendicular lines
Picture it
Example 3: through the point (-2, 3), parallel and perpendicular to y equals negative four x plus one.
Figure (svg): One point with two lines through it, one parallel to a given line and one perpendicular to it
Both new lines pass through the same point, so they differ only in slope. That is why the two answers share a great deal of their working.
Worked example
Example 3a. The slope is copied; the point is substituted.
\[ \text{Write the equation of the line through } (-2, 3) \text{ parallel to } y = -4x + 1. \]
Read the slope of the given line
Why: It is in slope-intercept form, so the coefficient of x is the slope.
\[ m 1 = -4 \]
Set the new slope equal to it
Why: Parallel lines have equal slopes, so nothing is changed.
\[ m 2 = -4 \]
Substitute into point-slope form with the given point
Why: Minus negative two becomes plus two inside the bracket.
\[ y - 3 = -4(x + 2) \]
Distribute and simplify
Why: Negative four times two is negative eight, and negative eight plus three is negative five.
\[ y = -4 x - 5 \]
Figure (svg): The solution to Worked example the parallel line shown as a ladder of expressions, one row per algebraic move
\[ y = -4x - 5 \]
Verify: check the point and compare the slopes
Why: At x equal to negative two: negative four times negative two is eight, minus five is three — the given y coordinate. And the coefficient of x matches the given line's, so the two really are parallel. Note the intercepts differ, negative five against one, which confirms they are distinct lines rather than the same one.
McDougal Littell Algebra 2 (Texas Edition), Ch. 2 Linear Equations and Functions — Lesson 2.4 Write Equations of Lines §2.4, pp. 99-99
Discrimination
You are told the given line's slope and what relationship is wanted.
Sort into buckets
Sort each case by what happens to the slope.
Worked example
Example 3b. Same point, and the slope is transformed rather than copied.
\[ \text{Write the equation of the line through } (-2, 3) \text{ perpendicular to } y = -4x + 1. \]
Take the negative reciprocal of the given slope
Why: The reciprocal of negative four is negative one quarter, and its negative is positive one quarter. Both operations, every time.
\[ m 2 = \frac{1}{4} \]
Substitute into point-slope form
Why: One quarter for m, and the bracket becomes x plus two as before.
\[ y - 3 = (\frac{1}{4}) (x + 2) \]
Distribute
Why: One quarter of x is x over four; one quarter of two is one half.
\[ y - 3 = (\frac{1}{4}) x + \frac{1}{2} \]
Add 3 to both sides
Why: One half plus three is seven halves.
\[ y = (\frac{1}{4}) x + \frac{7}{2} \]
Figure (svg): The solution to Worked example the perpendicular line shown as a ladder of expressions, one row per algebraic move
\[ y = \tfrac{1}{4}x + \tfrac{7}{2} \]
Verify: check the point and the product of the slopes
Why: At x equal to negative two: one quarter of negative two is negative one half, plus seven halves is three — the given y coordinate. And negative four times one quarter is negative one, which is the perpendicular condition from Lesson 2.2.
McDougal Littell Algebra 2 (Texas Edition), Ch. 2 Linear Equations and Functions — Lesson 2.4 Write Equations of Lines §2.4, pp. 99-99
Error analysis
A student writes the line through (4, -2) parallel to y equals 3x minus 1.
Annotate
On: \( m = 3 \;\Longrightarrow\; y = 3x - 1 \)
The given line contributes exactly one number, its slope. Everything else about the answer comes from the point.
Fill the middle
Guided Practice 5b: through (4, -2), perpendicular to y equals 3x minus 1.
Fill in the blanks
m_1 = 3 \;\Longrightarrow\; m_2 = -\frac{1}{3} \;\Longrightarrow\; y = -\tfrac______x - \tfrac______
Why: The reciprocal of 3 is one third, and its negative is negative one third. Substituting into point-slope form gives y plus 2 equals negative one third times x minus 4, which distributes to y plus 2 equals negative one third x plus four thirds, and subtracting 2 gives negative two thirds as the intercept. Checking the point: negative one third of four is negative four thirds, minus two thirds is negative two.
Comparison
Fill the blanks. Everything except the slope is shared.
Comparison matrix
| Step | Parallel through (-2, 3) | Perpendicular through (-2, 3) |
|---|---|---|
| Given line | y = -4x + 1 | y = -4x + 1 |
| New slope | -4 | 1/4 |
| Point-slope line | y - 3 = -4(x + 2) | y - 3 = (1/4)(x + 2) |
| Final equation | y = -4x - 5 | y = (1/4)x + 7/2 |
The two columns differ in exactly one place, the second row. Once the slope is settled, the two problems are identical work.
Counterexample
A classmate offers a shortcut for perpendicular lines.
\[ \text{to get a perpendicular line, just change the sign of the slope} \]
Discussion prompt
Take the line y equals 2x plus 1 and apply their shortcut. Show with the product test that the result is not perpendicular, and say what half of the operation they left out.
Hint: Compute the product of the two slopes.
Answer:
\[ m_1 = 2, \quad m_2 = -2 \;\Longrightarrow\; m_1 m_2 = -4 \neq -1 \]
Changing the sign alone gives a line that is a mirror image, not a perpendicular one — the two cross at a shallow angle. The missing half is inverting the fraction: the correct partner slope is negative one half, and 2 times negative one half is exactly negative one.
Section
Section 4
Concept
Two points determine a line, but neither of them is directly usable until you have a slope. Compute it with the slope formula, then feed it and either point into point-slope form. Both points give the same final equation.
\[ m = \frac{y_2 - y_1}{x_2 - x_1}, \quad \text{then } y - y_1 = m(x - x_1) \]
Using the other point is not just an alternative; it is the best available check, because an arithmetic slip almost never produces the same wrong answer twice.
Figure (svg): Two plotted points with the slope computed between them and the resulting line drawn through both
McDougal Littell Algebra 2 (Texas Edition), Ch. 2 Linear Equations and Functions — Lesson 2.4 Write Equations of Lines §2.4, pp. 100-100 — Write an equation given two points
Picture it
Example 4: the line through (5, -2) and (2, 10).
Figure (svg): Two plotted points with the slope computed between them and the resulting line drawn through both
The slope came out negative four, which the picture confirms: moving from the left point to the right one, the line drops steeply.
Worked example
Example 4. Slope first, then point-slope.
\[ \text{Write the equation of the line through } (5, -2) \text{ and } (2, 10). \]
Compute the slope with a consistent order
Why: Ten minus negative two is twelve on top; two minus five is negative three on the bottom.
\[ m = \frac{12}{-3} = -4 \]
Choose one of the two points
Why: Either works. Taking (2,10) keeps the numbers small.
\[ \text{use } (2, 10) \]
Substitute into point-slope form
Why: Negative four for m, two for x sub one, ten for y sub one.
\[ y - 10 = -4(x - 2) \]
Distribute and simplify
Why: Negative four times negative two is positive eight, and eight plus ten is eighteen.
\[ y = -4 x + 18 \]
Figure (svg): The solution to Worked example the line through two points shown as a ladder of expressions, one row per algebraic move
\[ y = -4x + 18 \]
Verify: redo it with the OTHER point
Why: Using (5,-2): y plus 2 equals negative four times x minus 5, which is negative 4x plus 20, so y equals negative 4x plus 18 — the same equation. Reaching the same answer from a different starting point is a much stronger check than re-reading the first calculation.
McDougal Littell Algebra 2 (Texas Edition), Ch. 2 Linear Equations and Functions — Lesson 2.4 Write Equations of Lines §2.4, pp. 100-100
Ranking
Writing the equation of a line through two given points.
Put in order
Why: Labelling first prevents the mixed-order error from Lesson 2.2. The slope must be found before point-slope form can be used, because that form has no slot for a second point. And the check uses the unused point deliberately: the point you substituted is guaranteed to work, so it proves nothing, while the other one tests the whole chain.
Worked example
Same procedure when the numbers do not divide evenly.
\[ \text{Write the equation of the line through } (-4, 9) \text{ and } (-8, 3). \]
Compute the slope
Why: Three minus nine is negative six; negative eight minus negative four is negative four. Two negatives give a positive.
\[ m = -6 / - 4 = \frac{3}{2} \]
Substitute with one of the points
Why: Taking (-4, 9), the bracket is x minus negative four, which is x plus four.
\[ y - 9 = (\frac{3}{2}) (x + 4) \]
Distribute
Why: Three halves of x is 3x over 2; three halves of four is six.
\[ y - 9 = (\frac{3}{2}) x + 6 \]
Add 9 to both sides
Why: Six plus nine is fifteen.
\[ y = (\frac{3}{2}) x + 15 \]
Figure (svg): The solution to Worked example two points with a fractional slope shown as a ladder of expressions, one row per algebraic move
\[ y = \tfrac{3}{2}x + 15 \]
Verify: substitute the OTHER point
Why: At x equal to negative eight: three halves of negative eight is negative twelve, plus fifteen is three — which is the second point's y coordinate. Both given points satisfy the equation, which is what it means for a line to pass through them.
McDougal Littell Algebra 2 (Texas Edition), Ch. 2 Linear Equations and Functions — Lesson 2.4 Write Equations of Lines §2.4, pp. 100-100
Trap
\[ \text{Through } (5, -2) \text{ and } (2, 10). \]
Compute the slope, then write only its size into the form
Why: The magnitude is carried forward and the sign is dropped somewhere in between.
\[ m = -4 \quad \text{but} \quad y - 10 = 4(x - 2) \]
\[ y = 4x + 2 \quad \text{(wrong)} \]
Substituting (5,-2) gives 22, not negative two, so the second point is nowhere near this line.
\[ \text{Through } (5, -2) \text{ and } (2, 10). \]
Carry the sign into the substitution
Why: The slope is negative four, and negative four is what goes into the bracket's multiplier.
\[ y - 10 = -4(x - 2) \;\Longrightarrow\; y = -4x + 18 \]
Both given points satisfy this, and a quick sketch agrees: from (2,10) to (5,-2) the line falls, so the slope must be negative.
Prediction
Commit before computing.
Predict first
You write the equation through two points using the first point, then again using the second. What happens?
Correct: The two equations are identical.
\[ y - 10 = -4(x - 2) \;\longrightarrow\; y = -4x + 18 \]
\[ y + 2 = -4(x - 5) \;\longrightarrow\; y = -4x + 18 \]
Why: Both points lie on the same line, and a line has exactly one equation in slope-intercept form. The intermediate point-slope lines look different — y minus 10 equals negative four times x minus 2 against y plus 2 equals negative four times x minus 5 — but they simplify to the same thing. That is exactly what makes redoing it with the other point a genuine check rather than a repetition.
Elimination
The line through (-4, 9) and (-8, 3).
Eliminate the wrong options
Which equation passes through both points?
Survives elimination: A
Why: The slope is 3 halves, and substituting either point into point-slope form gives an intercept of 15. Checking both: at x equal to -4, three halves of -4 is -6, plus 15 is 9; at x equal to -8, three halves of -8 is -12, plus 15 is 3. Both given points satisfy it.
Edge cases
The method assumes the slope formula produces a number.
Discussion prompt
What happens to the two-point method when the points are (3, 1) and (3, 7)? Work through the slope formula and say what goes wrong, then write the equation of the line anyway by a different route.
Hint: Compute the run before anything else.
Answer:
\[ m = \frac{7 - 1}{3 - 3} = \frac{6}{0} \quad \text{undefined} \]
The run is zero, so there is no slope and point-slope form cannot be used at all — it has an m-shaped hole in it that nothing fits.
The line still exists: both points have x equal to 3, so every point on the line does, and the equation is simply x = 3. This is the vertical-line case from Lesson 2.3, and it is the one line in the plane that this lesson's three methods cannot produce.
Section
Section 5
Concept
Real data arrives as two measurements. Choosing x to count years since the first measurement makes that first value the y-intercept, so the model's constant term is a number you actually observed rather than one extrapolated backwards.
\[ y = 0.086x + 2.00 \]
Had x been the calendar year itself, the intercept would be the value in year zero — a meaningless extrapolation two thousand years back.
Figure (svg): Two data points ten years apart, with the rate of change computed and turned into a linear model
McDougal Littell Algebra 2 (Texas Edition), Ch. 2 Linear Equations and Functions — Lesson 2.4 Write Equations of Lines §2.4, pp. 100-100 — Write a model using slope-intercept form
Picture it
Example 5: 2.00 million in 1993, rising to 2.86 million in 2003.
Figure (svg): Two data points ten years apart, with the rate of change computed and turned into a linear model
Defining x as years since 1993 puts the first data point at x equal to zero, which is exactly where the y-intercept lives. That is the whole reason for step one.
Worked example
Example 5, all three steps.
\[ \text{2.00 million in 1993 and } 2.86 \text{ million in 2003. Write a linear model.} \]
Define the variables explicitly
Why: Let x be the years since 1993 and y the participants in millions. Writing this down is step one because everything after it depends on the choice.
\[ x =\text{ years since } 1993 \]
Identify the initial value
Why: At x equal to zero the year is 1993 and the value is 2.00, so that is the y-intercept.
\[ b = 2.00 \]
Compute the rate of change
Why: The two points are (0, 2.00) and (10, 2.86), so the slope is 0.86 over 10.
\[ m = 0.086 \]
Write the verbal model, then the equation
Why: Participants equals initial number plus rate of change times years since 1993.
\[ y = 0.086 x + 2.00 \]
Figure (svg): The solution to Worked example participation in high school sports shown as a ladder of expressions, one row per algebraic move
\[ y = 0.086x + 2.00 \]
Verify: evaluate the model at both data points
Why: At x equal to 0 it gives 2.00 million, the 1993 figure. At x equal to 10 it gives 0.86 plus 2.00, which is 2.86 million, the 2003 figure. A two-point model must reproduce both points exactly, and this one does — which is the minimum any such model owes you.
McDougal Littell Algebra 2 (Texas Edition), Ch. 2 Linear Equations and Functions — Lesson 2.4 Write Equations of Lines §2.4, pp. 100-100
Real world
A subscription service had 1.4 million users in 2018 and 3.2 million in 2024.
Discussion prompt
Define your variables, write the linear model, and say what each of its two numbers means. Then use it to predict 2027 and say one reason the prediction might be wrong.
Hint: Let x count years since 2018 so the intercept is a measured value.
Answer:
\[ m = \frac{3.2 - 1.4}{6} = 0.3 \;\Longrightarrow\; y = 0.3x + 1.4 \]
The intercept 1.4 is the user count in millions in 2018; the slope 0.3 is the growth in millions of users per year. At x equal to 9, which is 2027, the model predicts 4.1 million.
It could be wrong because subscription growth is usually proportional rather than constant early on and then flattens as a market saturates — neither of which a straight line captures. Chapter 7's exponential models exist for exactly this.
Worked example
The same three steps applied to a decreasing quantity.
\[ \text{A tank held } 480 \text{ L in week } 0 \text{ and } 300 \text{ L in week } 6. \text{ Write a linear model.} \]
Define the variables
Why: Let x be the number of weeks since the first reading and y the volume in litres.
\[ x =\text{ weeks since the start} \]
Identify the initial value
Why: At x equal to zero the volume was 480 litres.
\[ b = 480 \]
Compute the rate of change
Why: Three hundred minus 480 is negative 180, over 6 weeks.
\[ m = -30 \]
Write the model and read its sign
Why: Negative thirty litres per week: the negative sign is the loss.
\[ y = -30 x + 480 \]
Figure (svg): The solution to Worked example a falling model from two measurements shown as a ladder of expressions, one row per algebraic move
\[ y = -30x + 480 \]
Verify: evaluate at both readings and find where it hits zero
Why: At x equal to 0 it gives 480 litres and at x equal to 6 it gives 480 minus 180, which is 300 — both readings reproduced. Setting y to zero gives x equal to 16, so the model predicts the tank empties in week 16, which is also where its useful domain ends.
Error analysis
A student models the sports data using the year itself as the input.
Annotate
On: \( \text{points } (1993, 2.00) \text{ and } (2003, 2.86) \;\Longrightarrow\; y = 0.086x - 169.2 \)
Both models are arithmetically valid. Step one of the procedure exists to choose the one whose constant term you can explain to someone.
Matching
Each model came from two real measurements.
Match the pairs
Why: In every case the intercept is the value at the moment x equals zero, which the variable definition chose, and the slope is the change per unit of x with its sign carrying the direction. The one falling model has a negative slope, and reading that sign as loss rather than as an error is part of interpreting a model.
Estimation
The sports model, y equals 0.086x plus 2.00, with x years since 1993.
Predict first
Roughly what does the model predict for 2013?
Correct: About 3.7 million.
\[ y = 0.086(20) + 2.00 = 1.72 + 2.00 = 3.72 \]
Why: Two thousand and thirteen is twenty years after 1993, and 0.086 per year for twenty years is about 1.7 million of growth, on top of the 2.00 million starting value. The exact value is 3.72 million. Estimating first catches the common slip of using the calendar year as x, which would give a wildly negative answer instead.
Socratic
One question, and nothing else on this slide.
\[ y = 0.086x + 2.00 \]
Discussion prompt
This model was built from exactly two measurements, ten years apart. What does it assume about the eight years in between, and what would you want to see before trusting it to describe them? Would a third data point strengthen the model, weaken it, or neither?
Hint: Think about what could have happened between 1993 and 2003 that two endpoints would hide.
Answer:
It assumes the change was steady throughout. Participation could have jumped in 1995 and flattened afterwards, or dipped and recovered, and the two endpoints would look identical either way.
What you would want is the intermediate years plotted, to see whether they fall near the line. A third point that lies close strengthens confidence; one that lies far off shows the linear assumption was wrong — so it can do either, and that is exactly why it is worth having.
Lesson 2.6 handles the realistic case, where many points are given and none of them lies exactly on any line.
Comparison
Fill the blanks. What you are given picks the row.
Comparison matrix
| Given | Form to use | First move |
|---|---|---|
| slope and y-intercept | y = mx + b | substitute both numbers |
| slope and a point | y - y1 = m(x - x1) | substitute, then distribute |
| two points | slope formula, then point-slope | compute the slope |
| a line and a point, parallel | point-slope | copy the given slope |
| a line and a point, perpendicular | point-slope | take the negative reciprocal |
Four of the five rows end in point-slope form. Learning that one form well covers almost the whole lesson.
Pattern
One routine writes any linear equation you will be asked for.
Step five is the only step that can catch a sign error made in step four, because it is the only one that returns to the information you were given.
OpenStax Algebra and Trigonometry 2e, §4.1 Linear Functions §4.1
Check
Slope and a point. Watch the bracket.
Check your understanding
Write the equation of the line through (-1, 6) with slope 4.
Answer: A
Why: Point-slope gives y - 6 = 4(x - (-1)), which is 4(x + 1), so y - 6 = 4x + 4 and y = 4x + 10. Checking: at x = -1, 4 times -1 plus 10 is 6.
Check
Perpendicular through a point. Both operations on the slope.
Check your understanding
Write the equation of the line through (4, -2) perpendicular to y = 3x - 1.
Answer: A
Why: The negative reciprocal of 3 is -1/3. Point-slope gives y + 2 = -(1/3)(x - 4), which is -(1/3)x + 4/3, so y = -(1/3)x - 2/3. Checking: at x = 4, -4/3 - 2/3 is -2.
Check
Two points. Find the slope first.
Check your understanding
Write the equation of the line through (5, -2) and (2, 10).
Answer: A
Why: The slope is (10 - (-2))/(2 - 5), which is 12 over -3, or -4. Using (2,10): y - 10 = -4(x - 2) gives y = -4x + 18. Both given points check.
Real world
A printing shop quotes 62 dollars for 200 flyers and 110 dollars for 400 flyers, and you suspect the price is a fixed setup fee plus a per-flyer cost.
Discussion prompt
Treat the two quotes as points, write the linear model, and say what each of its two numbers means to the shop. Then say what the model predicts for 1000 flyers, and one reason a real shop's pricing might not stay linear.
Hint: The quantity of flyers is the input; the price is the output.
Answer:
\[ m = \frac{110 - 62}{400 - 200} = \frac{48}{200} = 0.24 \]
\[ y - 62 = 0.24(x - 200) \;\Longrightarrow\; y = 0.24x + 14 \]
The 14 dollars is the setup fee — what you would pay for zero flyers, which is the plate and the labour — and 24 cents is the marginal cost per flyer. At 1000 flyers the model predicts 254 dollars.
Real pricing often has volume breaks: the per-flyer cost drops at 500 or 1000, which makes the graph a sequence of line segments rather than one line. A model built from two low-volume quotes will then over-predict at high volume.
Commit first
Answer, then rate your confidence honestly.
Predict first
How many different lines pass through the single point (2, 5)?
Correct: Infinitely many — one for every possible slope, plus the vertical one.
\[ y - 5 = m(x - 2) \quad \text{for every real } m, \; \text{plus } x = 2 \]
Why: A point fixes where a line is but says nothing about how steep it is, so every slope gives a different line through that point. That is exactly why every problem in this lesson supplies a second piece of information: a slope, another point, or a line to be parallel or perpendicular to. Two points, by contrast, determine exactly one line — which is why the two-point case has a unique answer.
Explain it
They can graph from y equals m x plus b but freeze when asked to write an equation from two points.
Discussion prompt
In four sentences or fewer, give them the whole procedure, explain why the slope has to come first, and tell them the check that catches almost every mistake.
Hint: The check involves the point they did not use.
Answer:
Compute the slope from the two points first, because the form you are about to use has a slot for a slope and no slot for a second point. Then put that slope and either one of the points into y minus y-one equals m times x minus x-one, and simplify to get y on its own.
The check is to substitute the OTHER point into your final equation. The point you used is guaranteed to fit, so it proves nothing; the one you did not use tests everything.
Exit ticket
Name the weakest spot before you close the deck.
Predict first
Which of these would you least want handed to you cold?
Correct: Whichever you picked is tonight's ten minutes, and each has a one-line fix.
Why: For negative coordinates, write x minus the coordinate in brackets before simplifying, so the double negative is visible. For perpendicular slopes, do both operations and then check the product is negative one. For the slope, label the points before subtracting anything. For models, define x as time since the first measurement so the intercept is a value you observed. Do five of your chosen kind rather than twenty mixed ones.
Connect it up
Paper. Fifteen minutes.
Draw it
Down the left of a page write the three cases: slope with intercept, slope with a point, and two points. Beside each write the form it uses and work one example of your own from start to finished slope-intercept equation, showing every line. In the middle of the page draw one coordinate plane, plot a single point, and draw four different lines through it, writing each of their equations beside them — this is the picture of why a point alone is never enough. In the right margin write a given line of your own and, through a point off it, work out both the parallel and the perpendicular equation, circling the one number that differs between the two calculations. At the bottom, take two real measurements you can look up, define your variables, build the linear model, and write one sentence saying what its intercept and its slope mean in words and units.
The circled number in the right margin should be the slope. If anything else differs between your two calculations, one of them has an error in it.
Recap
Five things, and the second one is the tool the other four are built on.
| If you are given | Then |
|---|---|
| Slope and y-intercept | Substitute into y = mx + b |
| Slope and any other point | Use point-slope form |
| Two points | Find the slope, then point-slope |
| A line to be parallel to | Copy its slope |
| A line to be perpendicular to | Invert and negate its slope |
Lesson 2.5 looks at the special case where the line passes through the origin, so the model has no constant term at all — direct variation.
McDougal Littell Algebra 2 (Texas Edition), Ch. 2 Linear Equations and Functions — Lesson 2.4 Write Equations of Lines §2.4, pp. 98-103 — everything on these slides traces back here
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