Parallel and perpendicular

INSTRUCTOR GUIDANCE EXAMPLE: Week Three Discussion

Parallel and Perpendicular

For this week’s discussion I am going to find the equations of lines that are parallel or

perpendicular to the given lines and which are passing through the specified point. First I

will work on the equation for the parallel line.

The equation I am given is y = -⅔ x + 2

The parallel line must pass through point (-6, -3)

I have learned that a line parallel to another line has the same slope as the other line, so

now I know that the slope of my parallel line will be -⅔. Since I now have both the

slope and an ordered pair on the line, I am going to use the point-slope form of a linear

equation to write my new equation.

y – y1 = m(x – x1) This is the general form of the point-slope equation

y – (-3) = -⅔[x – (-6)] Substituting in my known slope and ordered pair

y + 3 = -⅔x + (-⅔)6 Simplifying double negatives and distributing the slope

y = -⅔x – 4 – 3 Because (-⅔)6 = -4 and 3 is subtracted from both sides

y = -⅔x – 7 The equation of my parallel line!

This line falls as you go from left to right across the graph of it, the y-intercept is 7 units

below the origin, and the x-intercept is 10.5 units to the left of the origin.

Now I am ready to write the equation of the perpendicular line.

The equation I am given is y = -4x – 1

The perpendicular line must pass through point (0, 5)

I have learned that a line perpendicular to another line has a slope which is the negative

reciprocal of the slope of the other line so the first thing I must do is find the negative

reciprocal of –4.

The reciprocal of -4 is -¼ , and the negative of that is –(-¼) = ¼. Now I know my slope

is ¼ and my given point is (0, 5). Again I will use the point-slope form of a linear

equation to write my new equation.

y – y1 = m(x – x1)

y – 5 = ¼ (x – 0) Substituting in the slope and ordered pair

y – 5 = ¼ x The zero term disappears

y = ¼ x + 5 Adding 5 to both sides of the equation

The equation of my perpendicular line!

This line rises gently as you move from left to right across the graph. The y-intercept is

five units above the origin and the x-intercept is 20 units to the left of the origin.

[The answers to part d of the discussion will vary with students’ understanding.]