What are the three main forms of a linear equation?

Slope-intercept form: y = mx + b, where m is the slope and b is the y-intercept. Point-slope form: y − y₁ = m(x − x₁), where (x₁, y₁) is any point on the line and m is the slope. Standard form: Ax + By = C, where A, B, and C are integers with A ≥ 0. Slope-intercept is best for graphing; point-slope is best when you have a slope and a point; standard is best for finding x- and y-intercepts and for systems of equations.

How do you find the equation of a line from two points?

Step 1: compute the slope m = (y₂ − y₁)/(x₂ − x₁). Step 2: plug that slope and either one of the two points into point-slope form y − y₁ = m(x − x₁). Step 3 (optional): simplify to slope-intercept form y = mx + b by distributing and solving for y. It doesn't matter which of the two points you pick in step 2 — you get the same final line either way.

What is the difference between parallel and perpendicular slopes?

Two lines are parallel when their slopes are equal (m₁ = m₂). Two lines are perpendicular when their slopes are negative reciprocals of each other (m₁ · m₂ = −1). For example, a line with slope 2/3 is parallel to another line with slope 2/3, and perpendicular to a line with slope −3/2. Horizontal (m = 0) and vertical (m = undefined) lines are the special-case exception — they are perpendicular even though the product of their slopes isn't defined.

How do you write the equation of a horizontal or vertical line?

A horizontal line has slope 0 and every point on it has the same y-coordinate. Its equation is y = c, where c is that constant y-value. A vertical line has undefined slope and every point on it has the same x-coordinate. Its equation is x = c, where c is that constant x-value. You cannot write a vertical line in slope-intercept form because its slope is undefined.

When should you use point-slope form vs. slope-intercept form?

Use point-slope form when you know a slope and a point on the line — it lets you write the equation in one line without solving for the y-intercept first. Use slope-intercept form when you want to graph the line or when the problem gives you the y-intercept directly. In practice, most Algebra 1 problems ask you to write in point-slope first and then convert to slope-intercept as the final answer.

How is writing the equation of a line used in real life?

Any situation with a constant rate of change (linear relationship) can be modeled as y = mx + b. Examples: taxi fare (b = base charge, m = per-mile rate), phone plan (b = monthly fee, m = cost per GB), savings account with a fixed monthly deposit, distance-time for constant speed, and Celsius-to-Fahrenheit conversion (F = 1.8C + 32). Reading a word problem for the slope (rate) and y-intercept (starting value) is the core skill Algebra 1 tests over and over.

Where can I find a small-group Algebra 1 class in NYC?

SOMATH runs the Algebra 1 Ignite program (part of Young Fermats, for grades 6–9) at 226 W 79th St on the Upper West Side of NYC. Classes are small groups of 6–8 students taught by cofounder Marcelo Ambrozio (Northwestern-trained). It's a 48-class arc that covers everything from linear equations through quadratics, functions, and statistics. Call (646) 668-6151 or email hello@schoolofmath.us to book a free evaluation.

Young Fermats · Algebra 1 Ignite · Class 5 · Writing the Equation of a Line · Grades 6–9 · NYC Math Class

Writing the Equation of a Line — 25 Practice Questions with Theory & Hidden Answers (Algebra 1 Ignite Class 5)

A complete class-ready lesson on writing the equation of a line: slope-intercept form y = mx + b, point-slope form y − y₁ = m(x − x₁), writing from two points, horizontal and vertical lines, parallel and perpendicular slopes, and real-world modeling. Plus 25 practice questions with click-to-reveal step-by-step answers. Built for the SOMATH Algebra 1 Ignite class (Young Fermats, grades 6–9) on the Upper West Side of Manhattan.

2026-08-08 · By the SOMATH team · 226 W 79th St, UWS · (646) 668-6151

This is Class 5 of the SOMATH Algebra 1 Ignite arc. In Class 4 we learned how to read a line off a graph — slope, y-intercept, and how to plot it. Class 5 is the reverse direction: given the pieces of a line, write its equation. Every Algebra 1 midterm, every SAT, and every Regents exam tests some version of this skill.

Every question below has a hidden button that reveals the answer and the full reasoning, so a student can practice honestly and then check their thinking. Written by the same team that teaches Algebra 1 Ignite at SOMATH, a math-focused school on the Upper West Side of NYC run by cofounder Marcelo Ambrozio (Northwestern) and cofounder Vivianne Wright (Harvard).

How to use this in a class (50–60 min):

Warm-up: review slope-intercept form and the slope formula (10 min).
Work through the three main forms together on the board with one example each (15 min).
Students attempt the 25 questions with the answers hidden (25 min).
Reveal answers, re-teach any that missed 3+ students, and end on the modeling problem (10 min).

What’s in this lesson

Quick review: what slope and y-intercept mean
Slope-intercept form: y = mx + b
Point-slope form: y − y₁ = m(x − x₁)
Standard form: Ax + By = C
Writing the equation from two points
Horizontal and vertical lines
Parallel and perpendicular lines
Modeling real-world situations
25 practice questions
Answer key summary
About SOMATH & Algebra 1 Ignite

1. Quick review: what slope and y-intercept mean

The two ingredients of every line

Every non-vertical line is completely determined by two numbers:

Slope (m) — how much y changes for each 1-unit increase in x. Positive slope goes up, negative goes down, zero is flat.
y-intercept (b) — where the line crosses the y-axis. It’s the value of y when x = 0.

The slope formula from two points:

m = (y₂ − y₁) / (x₂ − x₁)

Rise over run. Order matters — subtract in the same direction on top and bottom.

2. Slope-intercept form: y = mx + b

The graphing-friendly form

The most common Algebra 1 form:

y = mx + b

where m is the slope and b is the y-intercept. If a problem gives you those two numbers directly, you’re done in one step — just plug them in.

Example

slope m = 3, y-intercept b = −4
→ y = 3x − 4

When to use this form: when you already know the y-intercept, when you need to graph the line, or as the final answer form for most Algebra 1 problems.

3. Point-slope form: y − y₁ = m(x − x₁)

The workhorse form

Point-slope form is the fastest way to write a line when you know a slope and any one point on the line.

y − y₁ = m(x − x₁)

where (x₁, y₁) is a known point and m is the slope. Just plug them in — no algebra needed for the “write it” part.

Example

slope m = 2, point (3, −1)
→ y − (−1) = 2(x − 3)
→ y + 1 = 2(x − 3) ← final in point-slope

Convert to slope-intercept:
y + 1 = 2x − 6
y = 2x − 7

When to use this form: any time you know a slope + a point but don’t know the y-intercept. It’s the fastest first step; you can always convert to y = mx + b at the end.

4. Standard form: Ax + By = C

The form for intercepts and systems

Standard form:

Ax + By = C

where A, B, and C are integers, and A ≥ 0. This form is useful for two reasons:

Finding intercepts fast: set x = 0 to get the y-intercept, set y = 0 to get the x-intercept.
Systems of equations: elimination is easiest when both equations are in standard form.

Example: convert to standard

y = −(2/3)x + 4
multiply both sides by 3: 3y = −2x + 12
add 2x to both sides: 2x + 3y = 12
→ standard form: 2x + 3y = 12

5. Writing the equation from two points

The three-step procedure

Given two points, the classic Algebra 1 problem. Three steps:

Slope: m = (y₂ − y₁) / (x₂ − x₁)
Plug into point-slope using either point.
Simplify to slope-intercept (distribute, then solve for y).

Example: through (1, 2) and (4, 11)

m = (11 − 2) / (4 − 1) = 9/3 = 3

Use point-slope with (1, 2):
y − 2 = 3(x − 1)
y − 2 = 3x − 3
y = 3x − 1

Sanity check: plug the other point (4, 11) into y = 3x − 1. 3(4) − 1 = 11. ✅ Both points satisfy the equation, so the line is right.

6. Horizontal and vertical lines

The two special cases every student misses

Line type	Slope	Equation	Example
Horizontal	m = 0	y = c	y = 4 (through (0, 4), (3, 4), etc.)
Vertical	undefined	x = c	x = −2 (through (−2, 0), (−2, 5), etc.)

Why vertical lines can’t use slope-intercept form. Slope-intercept requires a slope. A vertical line has an undefined slope (division by zero in the slope formula). So it lives outside y = mx + b and can only be written as x = c.

Trick to remember: horizontal lines have the letter that appears twice on top — they equal a constant y. Vertical lines equal a constant x.

7. Parallel and perpendicular lines

The slope relationships

Relationship	Slope rule	Example
Parallel	m₁ = m₂	y = 2x + 5 ∥ y = 2x − 3
Perpendicular	m₁ · m₂ = −1 (negative reciprocals)	y = 2x + 5 ⊥ y = −(1/2)x + 1

Negative reciprocal means: flip the fraction and change the sign. Slope 2/3 → perpendicular slope is −3/2. Slope −5 (= −5/1) → perpendicular slope is 1/5.

Example: line parallel to y = −3x + 4 through (2, 5)

Parallel ⇒ same slope: m = −3
Point-slope: y − 5 = −3(x − 2)
Distribute: y − 5 = −3x + 6
→ y = −3x + 11

Example: line perpendicular to y = (2/5)x + 1 through (0, −3)

Perpendicular slope = negative reciprocal of 2/5 = −5/2
y-intercept given: b = −3
→ y = −(5/2)x − 3

8. Modeling real-world situations

Reading the slope and y-intercept out of a word problem

Every “write the equation” word problem in Algebra 1 comes down to two questions:

What is the rate? → that’s the slope m. Look for phrases like per hour, per mile, each additional, every time.
What is the starting value? → that’s the y-intercept b. Look for phrases like base fee, initial amount, starts with, sign-up cost.

Example: taxi ride

“A taxi charges a $3.50 base fee plus $2.25 per mile.”
Starting value b = 3.50 (base fee)
Rate m = 2.25 (per-mile rate)
Let x = miles, y = total cost.
→ y = 2.25x + 3.50

25 practice questions

Read the question. Try the problem in your head or on scratch paper. Then click the button to reveal the answer and the full reasoning.

Question 1

Write the equation of a line in slope-intercept form with slope m = 5 and y-intercept b = −2.

Answer: y = 5x − 2

Plug directly into y = mx + b. m = 5 and b = −2 gives y = 5x + (−2) = 5x − 2.

Question 2

Write the equation of a line in slope-intercept form with slope m = −3/4 that passes through the y-axis at (0, 6).

Answer: y = −(3/4)x + 6

The y-intercept is where the line crosses the y-axis — that’s the point (0, 6), so b = 6. m is given as −3/4. So y = −(3/4)x + 6.

Question 3

Write the equation in point-slope form of a line with slope m = 2 that passes through the point (3, −5).

Answer: y + 5 = 2(x − 3)

Point-slope: y − y₁ = m(x − x₁). Plug in (x₁, y₁) = (3, −5) and m = 2:
y − (−5) = 2(x − 3) → y + 5 = 2(x − 3).

Watch the double negative: subtracting a negative flips to plus.

Question 4

Convert this point-slope equation to slope-intercept form: y − 4 = 3(x − 2)

Answer: y = 3x − 2

Distribute the 3, then isolate y.
y − 4 = 3x − 6
y = 3x − 6 + 4
y = 3x − 2

Question 5

Find the slope of the line through (2, 7) and (6, 15).

Answer: m = 2

Slope formula: m = (y₂ − y₁) / (x₂ − x₁) = (15 − 7) / (6 − 2) = 8/4 = 2.

Question 6

Write the equation in slope-intercept form of the line through (1, 2) and (4, 11).

Answer: y = 3x − 1

Step 1 — slope: m = (11 − 2)/(4 − 1) = 9/3 = 3.
Step 2 — point-slope using (1, 2): y − 2 = 3(x − 1).
Step 3 — simplify: y − 2 = 3x − 3 → y = 3x − 1.

Check the other point: 3(4) − 1 = 11. ✅

Question 7

Write the equation in slope-intercept form of the line through (−2, 5) and (3, −5).

Answer: y = −2x + 1

Slope: m = (−5 − 5)/(3 − (−2)) = −10/5 = −2.
Point-slope with (−2, 5): y − 5 = −2(x − (−2)) = −2(x + 2).
Distribute: y − 5 = −2x − 4 → y = −2x + 1.

Check with (3, −5): −2(3) + 1 = −5. ✅

Question 8

Write the equation of the horizontal line through (4, −7).

Answer: y = −7

A horizontal line has slope 0. Every point on it shares the same y-value. Since our point has y = −7, every other point on the line also has y = −7. Equation: y = −7.

Question 9

Write the equation of the vertical line through (−3, 8).

Answer: x = −3

A vertical line has undefined slope. Every point on it shares the same x-value. Our point has x = −3, so the line is x = −3. Note: this cannot be written in y = mx + b form — slope-intercept requires a defined slope.

Question 10

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

Answer: y = 4x − 5

Parallel ⇒ same slope. m = 4.
Point-slope with (2, 3): y − 3 = 4(x − 2).
Distribute: y − 3 = 4x − 8 → y = 4x − 5.

Question 11

Write the equation in slope-intercept form of the line perpendicular to y = 2x + 7 that passes through (4, 1).

Answer: y = −(1/2)x + 3

Perpendicular slope = negative reciprocal of 2 = −1/2.
Point-slope with (4, 1): y − 1 = −(1/2)(x − 4).
Distribute: y − 1 = −(1/2)x + 2 → y = −(1/2)x + 3.

Check perpendicularity: 2 · (−1/2) = −1. ✅

Question 12

A line passes through (0, −4) and has slope m = 2/3. Write its equation in slope-intercept form.

Answer: y = (2/3)x − 4

The point (0, −4) sits on the y-axis, so b = −4 — that’s the y-intercept. m = 2/3 is given. So y = (2/3)x − 4.

Shortcut: whenever the given point is on the y-axis, you can skip point-slope and plug straight into slope-intercept.

Question 13

Convert 2x + 3y = 12 to slope-intercept form. State the slope and y-intercept.

Answer: y = −(2/3)x + 4. Slope = −2/3, y-intercept = 4.

Solve for y.
2x + 3y = 12
3y = −2x + 12
y = −(2/3)x + 4

Slope m = −2/3, y-intercept b = 4.

Question 14

Convert y = −(3/4)x + 5 to standard form (Ax + By = C with integer coefficients and A ≥ 0).

Answer: 3x + 4y = 20

Multiply both sides by 4 to clear the fraction:
4y = −3x + 20
Add 3x to both sides:
3x + 4y = 20

A = 3 ≥ 0 ✅, all integers ✅.

Question 15

Find the x-intercept and y-intercept of 3x − 2y = 12.

Answer: x-intercept = (4, 0); y-intercept = (0, −6)

x-intercept: set y = 0. 3x = 12 → x = 4. Point: (4, 0).
y-intercept: set x = 0. −2y = 12 → y = −6. Point: (0, −6).

This is the main reason standard form exists — intercepts pop out in one step.

Question 16

Write the equation in slope-intercept form of the line through (−1, 4) and (3, −4).

Answer: y = −2x + 2

Slope: m = (−4 − 4)/(3 − (−1)) = −8/4 = −2.
Point-slope with (3, −4): y − (−4) = −2(x − 3) → y + 4 = −2x + 6 → y = −2x + 2.

Check with (−1, 4): −2(−1) + 2 = 4. ✅

Question 17

Which of these lines is parallel to y = −(3/5)x + 2?
A) y = (5/3)x + 2 B) y = −(3/5)x − 7 C) y = (3/5)x + 2 D) y = −(5/3)x + 2

Answer: B

Parallel ⇒ same slope. The original slope is −3/5. Only choice B has slope −3/5. Choice A has the negative reciprocal (5/3), so A is perpendicular, not parallel. C has the positive version, D has the wrong reciprocal.

Question 18

Write the equation in slope-intercept form of the line perpendicular to y = −(1/3)x + 2 that passes through the origin.

Answer: y = 3x

Perpendicular slope = negative reciprocal of −1/3 = 3.
Passes through the origin ⇒ b = 0.
So y = 3x + 0 = 3x.

Question 19

A line has a y-intercept of −3 and passes through (4, 5). Write its equation in slope-intercept form.

Answer: y = 2x − 3

The y-intercept gives us the point (0, −3). Now we have two points: (0, −3) and (4, 5).
Slope: m = (5 − (−3))/(4 − 0) = 8/4 = 2.
b = −3 (given).
→ y = 2x − 3.

Question 20

A line has an x-intercept of 6 and a y-intercept of −4. Write its equation in slope-intercept form.

Answer: y = (2/3)x − 4

The two intercepts give us two points: (6, 0) and (0, −4).
Slope: m = (−4 − 0)/(0 − 6) = −4/−6 = 2/3.
b = −4 (given).
→ y = (2/3)x − 4.

Question 21

Word problem: A gym charges a $40 sign-up fee plus $25 per month. Write an equation for the total cost y after x months.

Answer: y = 25x + 40

Starting value (y-intercept b): the $40 sign-up fee — the cost before any months have passed.
Rate (slope m): $25 per month — the cost that’s added for each additional month.
So y = 25x + 40.

Sanity check: at x = 0 months, cost = $40 (just the sign-up). At x = 3 months, cost = 25(3) + 40 = $115.

Question 22

Word problem: A tank starts with 500 gallons of water and drains at a constant rate. After 4 hours, the tank holds 380 gallons. Write an equation for the amount of water y (in gallons) after x hours.

Answer: y = −30x + 500

Starting value b: 500 gallons at time x = 0.
Rate m: the tank drained from 500 to 380 in 4 hours — that’s a loss of 120 gallons over 4 hours, or (380 − 500)/(4 − 0) = −30 gallons per hour.
So y = −30x + 500.

Sanity check: at x = 4, y = −30(4) + 500 = 380. ✅
Bonus: the tank is empty when y = 0. Solve: 0 = −30x + 500 → x = 500/30 ≈ 16.67 hours.

Question 23

Write the equation in slope-intercept form of the line perpendicular to 2x − 5y = 10 that passes through (−4, 3).

Answer: y = −(5/2)x − 7

Step 1 — find the original slope. Convert 2x − 5y = 10 to slope-intercept:
−5y = −2x + 10 → y = (2/5)x − 2. Original slope = 2/5.

Step 2 — perpendicular slope = negative reciprocal of 2/5 = −5/2.

Step 3 — point-slope with (−4, 3): y − 3 = −(5/2)(x − (−4)) = −(5/2)(x + 4).
Distribute: y − 3 = −(5/2)x − 10.
y = −(5/2)x − 7.

Question 24

Challenge: The points (2, k) and (5, 11) lie on a line with slope m = 3. Find k.

Answer: k = 2

Slope formula: 3 = (11 − k)/(5 − 2) = (11 − k)/3.
Multiply both sides by 3: 9 = 11 − k.
Solve: k = 11 − 9 = 2.

Sanity check: line through (2, 2) and (5, 11) ⇒ slope = (11 − 2)/(5 − 2) = 9/3 = 3. ✅

Question 25

Modeling challenge: A phone plan charges a $15 monthly fee plus $0.10 per minute of talk time. Last month, Ava’s bill was $47. Write an equation for the total bill y as a function of minutes x, then use it to find how many minutes Ava talked.

Answer: Equation y = 0.10x + 15; Ava talked 320 minutes.

Write the equation. Base fee = $15 (b), rate = $0.10 per minute (m). So y = 0.10x + 15.

Solve for minutes when y = 47.
47 = 0.10x + 15
32 = 0.10x
x = 320 minutes.

Sanity check: 320 min × $0.10 = $32, plus the $15 fee = $47. ✅

Why this problem matters: real Algebra 1 word problems combine writing the equation and using it to solve in a single question. Ignore either half and you lose the point.

Answer key summary

Q#	Answer	Q#	Answer	Q#	Answer
1	y = 5x − 2	10	y = 4x − 5	19	y = 2x − 3
2	y = −(3/4)x + 6	11	y = −(1/2)x + 3	20	y = (2/3)x − 4
3	y + 5 = 2(x − 3)	12	y = (2/3)x − 4	21	y = 25x + 40
4	y = 3x − 2	13	y = −(2/3)x + 4	22	y = −30x + 500
5	m = 2	14	3x + 4y = 20	23	y = −(5/2)x − 7
6	y = 3x − 1	15	(4, 0) & (0, −6)	24	k = 2
7	y = −2x + 1	16	y = −2x + 2	25	y = 0.10x + 15; 320 min
8	y = −7	17	B
9	x = −3	18	y = 3x

About SOMATH & Algebra 1 Ignite

SOMATH — School of Math is a math-focused school on the Upper West Side of Manhattan for students in grades 1–12. Algebra 1 Ignite is the flagship Young Fermats track (grades 6–9), a 48-class arc that covers everything from linear equations through quadratics, functions, sequences, and statistics — the full Regents Algebra 1 and pre-SAT scope. This post is Class 5 of that arc.

Classes are taught by cofounder Marcelo Ambrozio (Northwestern-trained, 15+ years teaching math in NYC) and the SOMATH team.

Location: 226 W 79th St, 1st Floor, New York, NY 10024 (Upper West Side)
Phone: (646) 668-6151
Email: hello@schoolofmath.us

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