Return to the Purplemath home page

 

Try a MathHelp.com demo lesson Join MathHelp.com Login to MathHelp.com

 

Index of free lessons | PM's free lessons in offline form | Forums | Search free lessons | Print this page | Local tutors

X

Straight-Line Equations:
   Parallel and Perpendicular Lines
(page 3 of 3)

Sections: Slope-intercept form, Point-slope form, Parallel and perpendicular lines


There is one other consideration for straight-line equations: finding parallel and perpendicular lines. Here is the usual format for the question:

  • Given the line 2x – 3y = 9 and the point (4, –1), find lines through the point that are:
    • (a) parallel to the given line and
      (b) perpendicular to it.

    In other words, they've given me a reference line — 2x – 3y = 9 — that I'll be comparing to, and some point somewhere else on the plane — namely, (4, –1). Then they want me to find the line through (4, –1) that is parallel to (that has the same slope as) 2x – 3y = 9. On top of that, they then want me to find the line through (4, –1) that is perpendicular to (that has a slope that is the negative reciprocal of the slope of) 2x – 3y = 9.

    Clearly, the first thing I need to do is solve "2x – 3y = 9" for "y=", so that I can find my reference slope: Copyright © Elizabeth Stapel 2000-2011 All Rights Reserved

      2x – 3y = 9
            –3y = –2x + 9
                y = ( 2/3)x – 3

    So the reference slope from the reference line is m = 2/3.

    Since a parallel line has an identical slope, then the parallel line through (4, –1) will have slope m = 2/3. Hey, now I have a point and a slope! So I'll use the point-slope form to find the line:

     

    ADVERTISEMENT

     

      y – (–1) = ( 2/3 )(x – 4)
      y + 1 = ( 2/3 ) x8/3
      y = ( 2/3 ) x8/33/3

      y = ( 2/3 ) x11/3

    This is the parallel line that they asked for.

    For the perpendicular line, I have to find the perpendicular slope. The reference slope is m = 2/3, and, for the perpendicular slope, I'll flip this slope and change the sign. Then the perpendicular slope is m = – 3/2. So now I can do the point-slope form. Note that the only change from the calculations I just did is that the slope is different now.

      y – (–1) = ( – 3/2 )(x – 4)
      y + 1 = ( – 3/2 ) x + 6

      y = ( – 3/2 ) x + 5

    Then the full solution to this exercise is:

      parallel:  y = ( 2/3 ) x11/3
      perpendicular:  y = ( – 3/2 ) x + 5


Warning: If a question asks you whether two given lines are "parallel, perpendicular, or neither", you must answer that question by finding their slopes, not by drawing a picture! Pictures can only give you a rough idea of what is going on, but you cannot tell "by looking" that lines with slopes of, say, m1 = 1.00 and m2 = 0.99 are NOT parallel, because they'll sure look parallel on their graphs. But since 1.00 does not equal 0.99, the lines are not parallel. Find the slopes; don't just draw the pictures.


You can use the Mathway widget below to practice finding a parallel line through a given point. Try the entered exercise, or type in your own exercise. Then click "Answer" to compare your answer to Mathway's. (The next widget is for finding perpendicular lines.)

(Clicking on "View Steps" on the widget's answer screen will take you to the Mathway site, where you can register for a free seven-day trial of the software.)


You can use the Mathway widget below to practice finding a perpendicular line through a given point. Try the entered exercise, or type in your own exercise. Then click "Answer" to compare your answer to Mathway's.

<< Previous  Top  |  1 | 2 | 3  |  Return to Index

Cite this article as:

Stapel, Elizabeth. "Straight-Line Equations: Parallel and Perpendicular Lines." Purplemath. Available from
    http://www.purplemath.com/modules/strtlneq3.htm. Accessed
 

 

 

FIND YOUR LESSON
This lesson may be printed out for your personal use.

Content copyright protected by Copyscape website plagiarism search

  Copyright 2000-2014  Elizabeth Stapel   |   About   |   Terms of Use   |   Linking   |   Site Licensing

 

 Feedback   |   Error?