Algebra Help: How to Find the Equation of a Line that is Parallel or Perpendicular to a Given Line

In my article Introduction to Linear Equations in Algebra I explain what linear equations are and how to find the equation of a line that passes through two points. Understanding and working with linear equations (or equations of lines) is a common skill the Algebra student will need to master. As a quick review, the standard form of a line is:

ax + by +c = 0 where a, b, and c are constants and a and b are not both zero.

The slope-intercept form of a line is:

y = mx + b

Where m and b are constants and m is the slope of the line and b is the y-intercept (that is, where the line crosses the y axis in the Cartesian coordinate system).

In this article I explain how to find the equation of a line that is either parallel or perpendicular to a given line and passes through a specific point.

How the Slopes of Parallel Lines are Related:
Two lines that are parallel have the same slope. That means if we know that two lines are parallel, the equations of the two lines would have the following slope-intercept forms:

Line #1: y = mx + b₁
Line #2: y = mx + b₂

The only difference between the two equations would be the y-intercepts (where the lines cross the y axis).

How the Slopes of Perpendicular Lines are Related:
Two lines that are perpendicular (make a right angle where they intersect) have slopes which are negative reciprocals of each other. That means if we know that two lines are perpendicular, the equations of the two lines would have the following slope-intercept forms:

Line #1: y = mx + b₁
Line #2: y = (-1/m)x + b₂

Let's now look at two sample problems that require the Algebra student to calculate the equation of a line that is parallel to a given line.

Algebra Problem #1: Find the Equation of a Parallel Line:
Find the equation of a line that is parallel to the line y = 3x - 2 and passes through the point (1, -2). (That is x = 1, y = -2, in the Cartesian coordinate system.)

Since we know that the slopes of parallel lines are equal, we see that the slope of our second line is 3. That gives us the equation for our second line as:

y = 3x + b

We now have to find our constant b (the y-intercept) for our second line. To do this, we need to substitute in the point we are given and solve for b.

-2 = 3(1) + b
-5 = b

Answer: Our equation of the parallel line is y = 3x - 5.

Algebra Problem #2: Find the Equation of a Parallel Line:
Find the equation of a line that is parallel to the line -20x + 2y - 10 = 0 and passes through the point (3, 4).

Before we can work with the original line we have to rewrite it in slope-intercept form.

2y = 20x + 10
y = 10x + 5

We now can do the same steps as in problem #2. The equation for our second line is:

y = 10x + b.

Substitute in our point and solve for b.

4 = 10(3) + b
-26 = b

Answer: Our equation of the parallel line is y = 10x - 26.

Let's modify our two sample problems slightly so that it requires the Algebra student to calculate the equation of a line that is perpendicular to a given line, instead of parallel.

Algebra Problem #3: Find the Equation of a Perpendicular Line:
Find the equation of a line that is perpendicular to the line y = 3x - 2 and passes through the point (1, -2).

Since we know that the slopes of perpendicular lines are negative reciprocals of each other, we see that the slope of our second line is (-1/3). That gives us the equation for our second line as:

y = (-1/3)x + b

We now have to find our constant b (the y-intercept) for our second line. To do this, we need to substitute in the point we are given and solve for b.

-2 = (-1/3)(1) + b
(-5/3) = b

Answer: Our equation of the perpendicular line is y = (-1/3)x - (5/3).

Algebra Problem #4: Find the Equation of a Perpendicular Line:
Find the equation of a line that is perpendicular to the line x + 2y - 10 = 0 and passes through the point (3, 4).

Before we can work with the original line we have to rewrite it in slope-intercept form.

2y = - x + 10
y = (-1/2)x + 5

Since we know that the slopes of perpendicular lines are negative reciprocals of each other, we see that the slope of our second line is 2. That gives us the equation for our second line as:

y = 2x + b.

Substitute in our point and solve for b.

4 = (2)(3) + b
-2 = b

Answer: Our equation of the perpendicular line is y = 2x - 2

Blessings!

Source
John Saxon, Jr. Saxon Algebra 2