Unit 3 · Lesson 4 of 5

Writing equations of lines (incl. parallel and perpendicular)

About 25 minutes · 5 steps · one idea: a slope and one point are all you need

You'll be able to: write the equation of a line in $y = mx + b$ form from a point and a slope, from two points, or from a parallel or perpendicular line.

1 Warm-up
2 The idea
3 Examples
4 Practice
5 Wrap-up

Step 1 of 5Warm-up

Two quick questions from earlier lessons. Answer from memory first. Pulling it out of your head is what makes it stick.

Write $4x + 2y = 6$ in slope-intercept form.

Subtract $4x$: $2y = -4x + 6$.

Divide every term by 2: $y = -2x + 3$.

Simplify $-3(x - 4) + 2$.

Distribute the $-3$ to both terms: $-3x + 12 + 2$.

Combine the numbers: $-3x + 14$.

Step 2 of 5The idea

A line is pinned down by its slope and any one point on it. The tool that joins them is point-slope form. You fill it in, then tidy it into $y = mx + b$.

Point-slope form. The line with slope $m$ through $(x_1, y_1)$ is \[y - y_1 = m(x - x_1)\]

Three steps. (1) Find $m$. (2) Put $m$ and the point into point-slope form. (3) Distribute and get $y$ alone.

Two points? Find $m = \dfrac{y_2 - y_1}{x_2 - x_1}$ first, then use either point.

Parallel lines have the same slope. Perpendicular lines have slopes that are negative reciprocals: flip the fraction and change the sign. Slope $\tfrac{2}{3}$ becomes $-\tfrac{3}{2}$; slope $4$ becomes $-\tfrac{1}{4}$.

Always finish with a quick check: put the point's $x$ into your answer and see that you get its $y$.

Why this shows up on the placement test: "find the equation of the line through these points" and "find the line perpendicular to this one" are classic placement questions.

Step 3 of 5Worked examples

Example 1: the line through $(-1, 5)$ and $(3, -3)$

Slope: $m = \dfrac{-3 - 5}{3 - (-1)} = \dfrac{-8}{4} = -2$.
Use the point $(-1, 5)$: $y - 5 = -2(x - (-1))$, which is $y - 5 = -2(x + 1)$.
Distribute: $y - 5 = -2x - 2$.
Add 5 to both sides: $y = -2x + 3$.
Check with the other point, $(3, -3)$: $-2(3) + 3 = -3$. It works, so the equation is right.

Example 2: perpendicular to $y = 2x + 1$, through $(4, 1)$

The given slope is $2 = \tfrac{2}{1}$. Flip it and change the sign: $m = -\tfrac{1}{2}$.
Point-slope: $y - 1 = -\tfrac{1}{2}(x - 4)$.
Distribute: $y - 1 = -\tfrac{1}{2}x + 2$. $-\tfrac{1}{2} \cdot (-4) = 2$.
Add 1: $y = -\tfrac{1}{2}x + 3$. Check: $-\tfrac{1}{2}(4) + 3 = 1$. Good.
For a parallel line through the same point, keep $m = 2$: $y - 1 = 2(x - 4)$, so $y = 2x - 7$.

The two lines meet at a right angle. The new line passes through $(4, 1)$.

Step 4 of 5Practice

Type your answer and press Check. If you miss twice, open the worked solution. Then mark how it went. Give each line in $y = mx + b$ form.

Write the equation of the line with slope $4$ through $(1, 3)$.

Point-slope: $y - 3 = 4(x - 1)$.

Distribute: $y - 3 = 4x - 4$. Add 3: $y = 4x - 1$.

Check: $4(1) - 1 = 3$.

Write the equation of the line with slope $-2$ through $(-3, 5)$.

Point-slope: $y - 5 = -2(x - (-3)) = -2(x + 3)$.

Distribute: $y - 5 = -2x - 6$. Add 5: $y = -2x - 1$.

Check: $-2(-3) - 1 = 5$.

Write the equation of the line through $(1, 2)$ and $(3, 8)$.

Slope: $m = \dfrac{8 - 2}{3 - 1} = \dfrac{6}{2} = 3$.

Point-slope with $(1, 2)$: $y - 2 = 3(x - 1)$, so $y - 2 = 3x - 3$ and $y = 3x - 1$.

Check with $(3, 8)$: $3(3) - 1 = 8$.

Write the equation of the line through $(-2, 4)$ and $(2, -2)$.

Slope: $m = \dfrac{-2 - 4}{2 - (-2)} = \dfrac{-6}{4} = -\dfrac{3}{2}$.

Point-slope with $(-2, 4)$: $y - 4 = -\tfrac{3}{2}(x + 2)$, so $y - 4 = -\tfrac{3}{2}x - 3$.

Add 4: $y = -\tfrac{3}{2}x + 1$. Check with $(2, -2)$: $-3 + 1 = -2$.

What is the slope of any line perpendicular to $y = -4x + 1$?

The given slope is $-4 = -\tfrac{4}{1}$.

Flip it to $-\tfrac{1}{4}$, then change the sign: $\tfrac{1}{4}$.

Check: $-4 \cdot \tfrac{1}{4} = -1$, as perpendicular slopes should.

Write the equation of the line parallel to $y = 3x - 2$ that passes through $(2, 1)$.

Parallel means the same slope: $m = 3$.

Point-slope: $y - 1 = 3(x - 2)$, so $y - 1 = 3x - 6$ and $y = 3x - 5$.

Check: $3(2) - 5 = 1$.

Write the equation of the line perpendicular to $2x + 3y = 6$ that passes through $(2, -1)$.

First find the given slope: $3y = -2x + 6$, so $y = -\tfrac{2}{3}x + 2$. Its slope is $-\tfrac{2}{3}$.

Perpendicular slope: flip and change the sign, $m = \tfrac{3}{2}$.

Point-slope: $y + 1 = \tfrac{3}{2}(x - 2)$, so $y + 1 = \tfrac{3}{2}x - 3$ and $y = \tfrac{3}{2}x - 4$.

Check: $\tfrac{3}{2}(2) - 4 = -1$.

A taxi charges a fixed fee plus a price per mile. A 2-mile ride costs $9 and a 6-mile ride costs $17. Which equation gives the cost $C$ in dollars for a ride of $m$ miles?

$C = 4m + 1$
$C = 2m + 9$
$C = 2m + 5$
$C = 5m + 2$

Use the points $(2, 9)$ and $(6, 17)$. Slope: $\dfrac{17 - 9}{6 - 2} = \dfrac{8}{4} = 2$ dollars per mile.

Point-slope: $C - 9 = 2(m - 2)$, so $C - 9 = 2m - 4$ and $C = 2m + 5$.

Check: $2(6) + 5 = 17$. Answer: (c).

Step 5 of 5Wrap-up

You can now write any line in $y = mx + b$ form: find the slope (same for parallel, flipped with the opposite sign for perpendicular), use $y - y_1 = m(x - x_1)$, then get $y$ alone.

Read or watch next: OpenStax Intermediate Algebra 2e, 3.3 Find the Equation of a Line (free textbook, with more practice at the end of the section).

Also good: Khan Academy: Writing slope-intercept equations (short videos). Quick rules: Lines and systems cheat sheet.

Something unclear? Ask your teacher bot a follow-up question, for example: "Why do perpendicular lines have slopes that multiply to -1?" It knows this course and can explain things another way.