Writing Equations of Lines

A linear equation describes a relationship with a constant rate of change. Its graph is a straight line.

There are several ways to write the equation of a line. The best method depends on the information you are given.

You might know:

• two points
• a slope and one point
• the slope and y-intercept
• a graph
• a table of values
• a real-world rate and starting value

The goal is to use the available information to determine an equation that describes every point on the line.

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Important Forms of a Linear Equation

Two forms are especially useful.

Slope-intercept form

y = mx + b

where:

• m = slope
• b = y-intercept

Point-slope form

y − y₁ = m(x − x₁)

where:

• m = slope
• (x₁, y₁) = known point on the line

Both forms can describe the same line.

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Choosing an Equation Form

The information given often suggests the easiest form.

Choosing an efficient form can greatly reduce the amount of algebra required.

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Writing an Equation from Two Points

Suppose a line passes through:

(2, 5) and (6, 13)

We do not initially know the slope or y-intercept.

First calculate the slope:

m = (y₂ − y₁)/(x₂ − x₁)

Substitute:

m = (13 − 5)/(6 − 2)

m = 8/4

m = 2

Now we know the slope.

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Using Point-Slope Form

Choose either known point.

Using:

(2, 5)

and:

m = 2

substitute into:

y − y₁ = m(x − x₁)

This gives:

y − 5 = 2(x − 2)

This is already a correct equation of the line.

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Converting to Slope-Intercept Form

Start with:

y − 5 = 2(x − 2)

Distribute:

y − 5 = 2x − 4

Add 5:

y = 2x + 1

Therefore, the same line can be written as:

Point-slope form: y − 5 = 2(x − 2)

Slope-intercept form: y = 2x + 1

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Checking the Equation

A useful habit is to check the equation using the second point.

Equation:

y = 2x + 1

Second point:

(6, 13)

Substitute:

13 = 2(6) + 1

13 = 12 + 1

13 = 13

The equation is correct.

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Either Point Can Be Used

Return to:

(2, 5) and (6, 13)

We found:

m = 2

Instead of using (2, 5), use:

(6, 13)

Point-slope form becomes:

y − 13 = 2(x − 6)

Expand:

y − 13 = 2x − 12

y = 2x + 1

Both points produce the same line.

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Worked Example: Negative Slope

Find the equation through:

(1, 8) and (5, 0)

First find slope:

m = (0 − 8)/(5 − 1)

m = −8/4

m = −2

Use (1, 8):

y − 8 = −2(x − 1)

Expand:

y − 8 = −2x + 2

Add 8:

y = −2x + 10

Check using (5, 0):

0 = −2(5) + 10

0 = 0

Correct.

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Worked Example with Negative Coordinates

Find the equation through:

(−2, 1) and (4, 13)

Slope:

m = (13 − 1)/(4 − (−2))

m = 12/6

m = 2

Use (−2, 1):

y − 1 = 2[x − (−2)]

y − 1 = 2(x + 2)

Expand:

y − 1 = 2x + 4

y = 2x + 5

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Writing an Equation from a Graph

A graph usually provides two pieces of information:

• slope
• y-intercept

If both can be identified, use:

y = mx + b

For example, suppose a graph crosses the y-axis at:

(0, 3)

and rises:

2 units for every 1 unit to the right

Then:

m = 2

b = 3

Therefore:

y = 2x + 3

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Finding Slope from a Graph

Choose two clear points on the line.

Suppose the points are:

(1, 3) and (4, 9)

Calculate:

m = (9 − 3)/(4 − 1)

m = 6/3

m = 2

If the graph crosses the y-axis at:

(0, 1)

then:

b = 1

Therefore:

y = 2x + 1

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Using Rise and Run from a Graph

Slope can also be found directly from a graph using:

m = rise/run

Suppose the line moves:

right 4

up 3

Then:

m = 3/4

If the y-intercept is:

(0, −2)

then:

y = (3/4)x − 2

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Graphs with Negative Slope

Suppose a graph crosses the y-axis at:

(0, 6)

and moves:

right 2

down 3

Then:

rise = −3

run = 2

Therefore:

m = −3/2

The equation is:

y = −(3/2)x + 6

The negative slope matches the fact that the line falls from left to right.

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When the y-Intercept Is Not Easy to Read

Sometimes the graph does not clearly show the y-intercept.

Suppose you can identify:

m = 3

and the point:

(4, 10)

Instead of trying to estimate the y-intercept, use point-slope form:

y − 10 = 3(x − 4)

Then simplify:

y − 10 = 3x − 12

y = 3x − 2

Now we know:

b = −2

This is often more accurate than estimating an intercept from a graph.

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Writing an Equation from a Point and Slope

Suppose you know:

m = 4

and the line passes through:

(3, 7)

Point-slope form is the natural choice:

y − 7 = 4(x − 3)

If slope-intercept form is required:

y − 7 = 4x − 12

y = 4x − 5

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Worked Example: Fractional Slope

A line has:

m = 2/3

and passes through:

(6, 5)

Use point-slope form:

y − 5 = 2/3(x − 6)

Distribute:

y − 5 = 2/3x − 4

Add 5:

y = 2/3x + 1

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Worked Example: Negative Point

A line has slope:

−3

and passes through:

(−2, 4)

Use point-slope form:

y − 4 = −3[x − (−2)]

Simplify:

y − 4 = −3(x + 2)

Expand:

y − 4 = −3x − 6

y = −3x − 2

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Finding b Directly

You can also find the y-intercept by substituting a point into:

y = mx + b

Suppose:

m = 3

and:

(4, 17)

Substitute:

17 = 3(4) + b

17 = 12 + b

b = 5

Therefore:

y = 3x + 5

This method is especially useful when the final answer must be in slope-intercept form.

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Point-Slope or Slope-Intercept?

Suppose you are given:

m = −2

point = (7, 5)

Point-slope form gives the equation immediately:

y − 5 = −2(x − 7)

If the question only asks for an equation, you may be finished.

If the question specifically asks for slope-intercept form, continue:

y − 5 = −2x + 14

y = −2x + 19

The most appropriate form depends on the purpose of the problem.

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Writing an Equation from a Table

Tables can also represent linear relationships.

Consider:

Each time x increases by 1:

y increases by 3.

Therefore:

m = 3

When x = 0:

y = 4

Therefore:

b = 4

The equation is:

y = 3x + 4

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A Table Without x = 0

Consider:

Calculate the slope using two rows:

m = (18 − 9)/(5 − 2)

m = 9/3

m = 3

Now use:

(2, 9)

in:

y = mx + b

9 = 3(2) + b

9 = 6 + b

b = 3

Therefore:

y = 3x + 3

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Checking for a Constant Rate of Change

Before writing a linear equation from data, make sure the relationship is actually linear.

Consider:

Each time:

Δx = 1

and:

Δy = 3

Therefore:

Δy/Δx = 3

throughout the table.

The rate of change is constant, so a linear model is appropriate.

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Unequal x-Intervals

Do not compare only the changes in y when x-values are unevenly spaced.

Consider:

First interval:

m = (11 − 5)/(3 − 1)

m = 6/2

m = 3

Second interval:

m = (23 − 11)/(7 − 3)

m = 12/4

m = 3

The rate of change remains constant.

Therefore, the data are linear.

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Special Case: Horizontal Lines

Suppose two points are:

(2, 5) and (8, 5)

Slope:

m = (5 − 5)/(8 − 2)

m = 0

Therefore, the line is horizontal.

Its equation is:

y = 5

This can also be written:

y = 0x + 5

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Special Case: Vertical Lines

Suppose two points are:

(4, 2) and (4, 9)

The x-coordinate remains:

4

The slope calculation would require division by zero.

Therefore, the slope is undefined.

The equation is:

x = 4

A vertical line cannot be written in ordinary slope-intercept form because y is not a function of x.

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Recognizing Which Variable Depends on the Other

In a real-world model, decide which variable is the input and which is the output.

Usually:

x = independent variable

y = dependent variable

For example:

hours worked → input

money earned → output

Then:

money earned depends on hours worked

The equation might be:

y = 18x + 25

where:

18 = dollars earned per hour

25 = starting amount

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Modeling a Starting Value and Rate

Many real-world linear situations follow:

output = rate × input + starting value

This has exactly the same structure as:

y = mx + b

where:

m = rate of change

b = initial value

This connection makes slope-intercept form particularly useful for modeling.

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Real-World Example: Taxi Fare

A taxi company charges:

$5 starting fee

plus:

$2.50 per kilometre

Let:

x = distance travelled in kilometres

y = total cost in dollars

Rate:

m = 2.50

Starting value:

b = 5

Therefore:

y = 2.5x + 5

Interpretation:

• slope = $2.50/km
• y-intercept = $5 initial charge

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Using the Taxi Model

Suppose the trip is:

12 km

Use:

y = 2.5x + 5

Substitute:

y = 2.5(12) + 5

y = 30 + 5

y = $35

The equation allows us to make predictions within the model.

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Real-World Example: Savings

A student has:

$120

and saves:

$35 each week

Let:

x = weeks

y = savings

Then:

m = 35

b = 120

Equation:

y = 35x + 120

After 8 weeks:

y = 35(8) + 120

y = 280 + 120

y = $400

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Real-World Example: Decreasing Quantity

A tank initially contains:

900 L

Water drains at:

25 L/min

Let:

x = time in minutes

y = water remaining

Because the amount decreases:

m = −25

Initial amount:

b = 900

Therefore:

y = −25x + 900

The negative slope represents the decrease.

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Modeling from a Known Measurement

Sometimes the initial value is unknown.

Suppose a plant grows at a constant modeled rate of:

2.5 cm/week

At week 4, its height is:

23 cm

Known point:

(4, 23)

Slope:

m = 2.5

Point-slope form is convenient:

y − 23 = 2.5(x − 4)

Simplify:

y − 23 = 2.5x − 10

y = 2.5x + 13

The model predicts an initial height of:

13 cm

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Real Measurements and Linear Models

Real data do not always fall exactly on a straight line.

For example, plant growth, temperature, sales, and travel may change at different rates.

A linear equation is appropriate when:

• the rate is approximately constant
• a straight line reasonably represents the data
• the model is used over an appropriate range

An equation may be mathematically valid for every x-value but unrealistic outside the context.

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Interpolation

Interpolation means using a model to estimate a value between known data values.

Suppose measurements were collected at:

x = 2 and x = 10

Using the model to estimate a value at:

x = 6

is interpolation.

This is often reasonably reliable when the linear relationship is appropriate.

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Extrapolation

Extrapolation means using a model to predict beyond the known data range.

If measurements were collected from:

x = 2 to x = 10

using the equation to predict:

x = 50

is extrapolation.

This can be less reliable because the relationship may not remain linear that far beyond the observed data.

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Comparing Different Equation Forms

Consider the line:

y = 3x − 4

The same line could be written using the point (2, 2):

y − 2 = 3(x − 2)

These equations look different but represent the same line.

Slope-intercept form makes it easy to see:

• slope
• y-intercept

Point-slope form makes it easy to see:

• slope
• a known point

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Standard Form

Another common linear form is:

Ax + By = C

For example:

2x + 3y = 12

This is called standard form.

To reveal the slope and y-intercept, rearrange:

3y = −2x + 12

Divide by 3:

y = −(2/3)x + 4

Now we can see:

m = −2/3

b = 4

Different equation forms emphasize different features of a line.

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Selecting the Most Appropriate Form

Suppose you know:

slope = 4

and:

y-intercept = −2

The easiest form is:

y = 4x − 2

Suppose instead you know:

slope = 4

and the line passes through:

(7, 10)

The easiest starting form is:

y − 10 = 4(x − 7)

Choosing the form that matches the available information reduces unnecessary calculations.

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Writing a Parallel Line

Suppose you need a line parallel to:

y = 3x + 5

passing through:

(4, 1)

Parallel lines have equal slopes.

Therefore:

m = 3

Use point-slope form:

y − 1 = 3(x − 4)

Simplify:

y − 1 = 3x − 12

y = 3x − 11

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Writing a Perpendicular Line

Suppose you need a line perpendicular to:

y = 2x + 4

passing through:

(6, 3)

Original slope:

m = 2

Perpendicular slope:

m = −1/2

Use point-slope form:

y − 3 = −1/2(x − 6)

Simplify:

y − 3 = −1/2x + 3

y = −1/2x + 6

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Determining Whether a Point Fits an Equation

Suppose the equation is:

y = 3x + 2

Does:

(4, 14)

lie on the line?

Substitute:

14 = 3(4) + 2

14 = 14

Yes.

Now test:

(5, 14)

14 = 3(5) + 2

14 = 17

False.

Therefore:

(5, 14) does not lie on the line.

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Finding a Missing Coordinate

Suppose:

y = 4x − 3

and a point on the line has:

x = 5

Then:

y = 4(5) − 3

y = 20 − 3

y = 17

Therefore:

(5, 17)

lies on the line.

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Finding x from an Equation

Suppose:

y = 3x + 5

and:

y = 26

Substitute:

26 = 3x + 5

21 = 3x

x = 7

Therefore:

(7, 26)

lies on the line.

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Checking an Equation Using Multiple Representations

A strong way to verify a linear equation is to compare several representations.

For example, suppose you find:

y = 2x + 3

Check:

Slope

Does the line rise 2 for every 1 right?

y-intercept

Does it cross at (0, 3)?

Known points

Do the given coordinates satisfy the equation?

Table

Do calculated values match the data?

If all of these agree, the equation is likely correct.

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A Complete Two-Point Problem

Find the equation through:

(−3, 8) and (5, −4)

First calculate slope:

m = [−4 − 8]/[5 − (−3)]

m = −12/8

m = −3/2

Use (−3, 8):

y − 8 = −3/2[x − (−3)]

y − 8 = −3/2(x + 3)

Convert:

y − 8 = −3/2x − 9/2

Add 8:

y = −3/2x − 9/2 + 16/2

y = −3/2x + 7/2

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Checking the Complete Example

Use:

(5, −4)

Equation:

y = −3/2x + 7/2

Substitute:

−4 = −3/2(5) + 7/2

−4 = −15/2 + 7/2

−4 = −8/2

−4 = −4

Correct.

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A Complete Real-World Model

A streaming service charges a fixed monthly fee of:

$12

plus:

$3 for each premium rental

Let:

x = number of premium rentals

y = total monthly cost

The fixed fee represents:

b = 12

The cost per rental represents:

m = 3

Therefore:

y = 3x + 12

For 6 rentals:

y = 3(6) + 12

y = 18 + 12

y = $30

The equation provides both a description of the relationship and a way to make predictions.

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Domain Restrictions in Context

The equation:

y = 3x + 12

mathematically allows values such as:

x = −4

or:

x = 2.7

But if x represents the number of premium rentals, those values may not make sense.

The realistic domain might be:

x = 0, 1, 2, 3, ...

Therefore, mathematical models must be interpreted according to their context.

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Discrete and Continuous Models

Some linear situations are continuous.

Examples:

• time
• distance
• temperature
• volume

Values between measurements may make sense.

Other situations are discrete.

Examples:

• number of tickets
• number of students
• number of products
• number of rentals

A line equation may describe the relationship algebraically, but only certain points may represent realistic possibilities.

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A Reliable Strategy: Two Points

When given two points:

1. Label the points.
2. Calculate slope using m = (y₂ − y₁)/(x₂ − x₁).
3. Simplify the slope.
4. Choose either point.
5. Substitute into point-slope form.
6. Write the equation.
7. Convert to slope-intercept form if required.
8. Check the equation using the other point.

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A Reliable Strategy: From a Graph

When given a graph:

1. Identify two clear points.
2. Calculate rise and run.
3. Determine the slope.
4. Find the y-intercept if visible.
5. Write y = mx + b.
6. If the y-intercept is unclear, use a known point and point-slope form.
7. Check that the slope sign matches the graph's direction.
8. Verify another point.

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A Reliable Strategy: Point and Slope

When given a point and slope:

1. Identify m.
2. Identify (x₁, y₁).
3. Write y − y₁ = m(x − x₁).
4. Substitute carefully.
5. Simplify double negatives.
6. Leave in point-slope form if appropriate.
7. Convert to y = mx + b if required.
8. Check the original point.

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A Reliable Strategy: Real-World Models

When writing a linear model:

1. Identify the input variable.
2. Identify the output variable.
3. Decide whether a linear model is appropriate.
4. Identify the rate of change.
5. Determine whether an initial value is known.
6. Choose a useful equation form.
7. Write the equation.
8. Include meaningful units.
9. Use the equation to calculate or predict values.
10. Interpret the result in context.
11. Consider whether the input is discrete or continuous.
12. Consider the realistic domain of the model.

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Common Mistakes

Finding b before finding the slope when given two points

Usually find the slope first.

Using the slope formula backward

Slope is:

change in y/change in x

Changing subtraction order halfway through

If you calculate y₂ − y₁, also calculate x₂ − x₁.

Forgetting negative coordinates

Use parentheses when substituting negative values.

Assuming the y-intercept must be visible

If it is not, use a known point and the slope.

Choosing slope-intercept form when point-slope form is much easier

Use the form that matches the information given.

Forgetting that horizontal lines have slope zero

Their equations have the form:

y = constant

Trying to write a vertical line as y = mx + b

Vertical lines have equations such as:

x = 4

Ignoring units in real-world problems

A slope such as 5 is much more meaningful as 5 km/h, $5/item, or 5 L/min.

Assuming every data set is linear

Check whether the rate of change is reasonably constant.

Using a model outside a realistic domain

A mathematically valid answer may not make sense physically.

Treating discrete data as continuous

If x represents people or products, fractional values may not be meaningful.

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Key Terms

Linear equation — equation whose graph is a straight line.

Slope — constant rate of change of a nonvertical linear relationship.

y-intercept — point where a graph crosses the y-axis.

Slope-intercept form — equation written as y = mx + b.

Point-slope form — equation written as y − y₁ = m(x − x₁).

Standard form — linear equation commonly written as Ax + By = C.

Rate of change — change in output compared with change in input.

Initial value — value of the output when the input is zero, when meaningful.

Linear model — linear equation used to represent a real relationship.

Interpolation — estimating within the range of known data.

Extrapolation — predicting beyond the range of known data.

Independent variable — input variable.

Dependent variable — output variable whose value depends on the input.

Domain — set of allowable input values.

Discrete data — values that occur as separate countable possibilities.

Continuous data — values that can vary through an interval.

Equivalent equations — different equations representing the same relationship or line.

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Key Takeaways

• A straight line can be represented by a linear equation.
• Different information may make different equation forms more convenient.
• When given two points, first calculate the slope.
• After finding the slope, point-slope form provides a direct way to write the equation.
• Either of two known points can be used in point-slope form.
• Slope-intercept form is useful when the slope and y-intercept are known.
• A graph can provide slope through rise/run and the y-intercept through its crossing of the y-axis.
• If the y-intercept cannot be read accurately, use a known point instead.
• A table represents a linear relationship when its rate of change is constant.
• Unequal x-intervals require comparing Δy/Δx rather than simply comparing changes in y.
• Horizontal lines have equations of the form y = constant.
• Vertical lines have equations of the form x = constant.
• Point-slope and slope-intercept forms can represent the same line.
• Real-world linear models often have the structure output = rate × input + starting value.
• Slope should be interpreted with units in real-world models.
• The y-intercept often represents an initial value, but x = 0 must make sense in context.
• Interpolation estimates values within known data.
• Extrapolation predicts beyond known data and can be less reliable.
• Real-world models should be used only over reasonable domains.
• Discrete and continuous variables should be distinguished.
• Substitution provides a useful way to verify an equation.
• Selecting the most appropriate equation form makes solving linear problems faster and less error-prone.

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Check Your Understanding

1. Describe when slope-intercept form is more convenient than point-slope form.
2. Find the equation of the line passing through (2, 4) and (6, 12).
3. Find the equation of the line passing through (−3, 2) and (5, 18).
4. A line has slope −4 and passes through (3, 7). Write its equation in point-slope form and slope-intercept form.
5. A graph crosses the y-axis at (0, −3) and rises 5 units for every 2 units moved right. Write its equation.
6. A line passes through (4, 9) and has slope 3/2. Determine its y-intercept and write its equation in slope-intercept form.
7. Determine the equation of the horizontal line passing through (5, −7).
8. Determine the equation of the vertical line passing through (−4, 6).
9. A table contains the points (2, 7), (5, 16), and (9, 28). Determine whether the relationship is linear. If it is, write its equation.
10. Explain why the equation of a line can be written in different forms without changing its graph.
11. A taxi charges a $6 starting fee and $2.25 per kilometre. Write an equation for the total cost y of travelling x kilometres. Explain the meaning of the slope and y-intercept.
12. A tank contains 700 L of water and loses 15 L each minute. Write a linear model and use it to determine the amount remaining after 20 minutes.
13. A quantity increases at 8 units per hour. After 5 hours its value is 62. Write a point-slope equation, convert it to slope-intercept form, and determine its modeled value at x = 0.
14. Explain why extrapolating a real-world linear model far beyond the range of measured data may produce unreliable predictions.
15. A student is given the points (−2, 5) and (4, 17) and writes y = 3x + 11. Determine whether the equation is correct. If it is not, identify the error and write the correct equation.