DWG NO. 03.4 — Lesson 4 of 4

Writing Equations of Lines in a Geometric Context

Unit 3: Parallel and Perpendicular Lines · ~15–30 min

Objective

Write the equation of a line, given a point it passes through and a requirement that it be parallel or perpendicular to another line.

Two forms worth knowing

Form	Equation	Best for
Slope-intercept	y = mx + b	Reading off slope (m) and y-intercept (b) at a glance
Point-slope	y − y₁ = m(x − x₁)	Building an equation from a slope and any one point (x₁, y₁)

Point-slope form is the workhorse for this lesson: once you know the slope you need and one point the line must pass through, it lets you write the equation immediately — then simplify to slope-intercept form if that's what's asked for.

The steps

Find the slope of the given line (rewrite it in slope-intercept form first if it isn't already).
If the new line must be parallel, use that same slope. If it must be perpendicular, use the negative reciprocal.
Substitute the slope and the given point into point-slope form.
Distribute and simplify to slope-intercept form.

The new line passes through P and is drawn parallel to g — it keeps g's slope but shifts to hit a different point.

Worked Example 1 · Parallel through a point

ProblemWrite the equation of the line parallel to y = (2/3)x − 1 that passes through (3, 5).

1.The given line has slope 2/3. A parallel line keeps the same slope: m = 2/3.

2.Substitute into point-slope form using (3, 5): y − 5 = (2/3)(x − 3).

3.Distribute and simplify: y − 5 = (2/3)x − 2, so y = (2/3)x + 3.

y = (2/3)x + 3

Worked Example 2 · Perpendicular through a point

ProblemWrite the equation of the line perpendicular to y = (2/3)x − 1 that passes through (4, −1).

1.The given line has slope 2/3. A perpendicular line uses the negative reciprocal: m = −3/2.

2.Substitute into point-slope form using (4, −1): y − (−1) = (−3/2)(x − 4).

3.Distribute and simplify: y + 1 = (−3/2)x + 6, so y = (−3/2)x + 5.

y = (−3/2)x + 5

Guided practice

1.Write the equation of the line parallel to y = 4x + 7 that passes through (1, 2).
Show answer
y − 2 = 4(x − 1), which simplifies to y = 4x − 2.
2.Write the equation of the line perpendicular to y = 4x + 7 that passes through (8, 1).
Show answer
The perpendicular slope is −1/4. y − 1 = (−1/4)(x − 8), which simplifies to y = (−1/4)x + 3.
3.A line passes through (0, −3) and (2, 1). Write the equation of the line parallel to it through (4, 0).
Show answer
Slope of the given line: (1 − (−3)) ÷ (2 − 0) = 2. Parallel line: y − 0 = 2(x − 4), which simplifies to y = 2x − 8.
4.Write the equation of the line perpendicular to x = 5 (a vertical line) that passes through (2, 6).
Show answer
A line perpendicular to a vertical line is horizontal, with slope 0. Through (2, 6), that's y = 6.
5.Explain why you don't need to know the y-intercept of the original line to find a parallel or perpendicular equation through a new point.
Show answer
Parallel and perpendicular relationships depend only on slope, not on where a line crosses the y-axis. Once you have the right slope and the new point, point-slope form builds the whole equation without ever needing the original line's intercept.

In-person Review & Practice Day 3Not a web page — work problems, ask questions, get feedback in class.

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