1 / 6

Solving Equations of Parallel and Perpendicular lines

Solving Equations of Parallel and Perpendicular lines. The following examples will help you to work through problems involving Parallel and Perpendicular lines .

chadwick-morse
Download Presentation

Solving Equations of Parallel and Perpendicular lines

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Solving Equations of Parallel and Perpendicular lines • The following examples will help you to work through problems involving Parallel and Perpendicular lines. • Example 1 will show you how to convert a line in Standard Form to a line in Slope Intercept form using Algebraic methods. Then using your result, finding the equation of the line that is perpendicluar to the original given line. • It is the least efficient method

  2. Ax + By = C 2x + 4y = 20 Subract 2x from both sides 4y = -2x + 20 Divide everything by 4y = -2/4x + 20/4 y = -1/2x + 5 Example 1Write the equation of a line in slope interceptform that passes through the point (3,-3), and is perpendicular to the following line: 2x + 4y = 20 • In this example we will convert standard form to slope intercept form. Then we will find the slope of the perpendiular line then use the point slope formand the given point to find the equation we are looking for. • First off….notice that the question asks for the final answer to be in slope interceptform. y = mx + b

  3. We have just converted the line:2x +4y = 20 from standard form to slope intercept form: y = -1/2x +5 And the slope of our original line is:m=-1/2 We understand that slopes of perpendicuar lines have the relationship of being Opposite reciprocals So the slope of the line that is perpendicular to our original line must be m = 2 Standard Form 2x + 4y = 20 Converted to Slope Intercept Form: y = -1/2x +5 Perpendicular lineshave opposite reciprocal slope: m = 2 The next step is to find the equation of the line we are looking for. Write the equation of a line in slope interceptform that passes through the point (3,-3), and is perpendicular to the following line: 2x + 4y = 20

  4. We now have the equation of the original line in slope intercept form:y = -1/2x +5 We also know that the slope of our new (perpendicular) line must be the opposite reciprocal of our original line so: The opposite reciprocal Of “-1/2” is “2”, thus the slope of Our new line must be 2 We now use the Point Slope Form “MOM”and the point that was given to us (3, -3), and substitute our information then simplify. y - y1 = m(x - x1) The y value of our point is-3 y – (-3) = m(x - x1) The x value of our point is3 y – (-3) = m(x - 3) The slope of our perpendicular line is 2 y – (-3) = 2(x - 3) Example 1Write the equation of a line in slope interceptform that passes through the point (3,-3), and is perpendicular to the following line: 2x + 4y = 20

  5. We are now ready to calculate the equation of the line that is perpendicular to our original line that passes through the given point. To recap: our original line: 2x + 4y = 20 Converted to Slope intercept form:y = -1/2x +5 Finding the slope of the new line: m = 2 Using the given point values: ( 3 , -3 ) Substituting the information into the Point Slope Form y – (-3) = 2(x – 3) y – (-3) = 2(x – 3) Distributing: y + 3 = 2x – 6 Simplifying: y = 2x – 6 – 3 Answer: y = 2x – 9 This is the equation of the line that is perpendicular to:2x + 4y = 20 That passes through the point( 3, -3) UR Done!!! Example 1Write the equation of a line in slope interceptform that passes through the point (3,-3), and is perpendicular to the following line: 2x + 4y = 20

  6. Try the following examples and email me your answer.nfmath@yahoo.com • Write the equation of a line in slope intercept form that passes through the point (3,-3), and is perpendicular to the following line: • 1.10x + 2y = 30 2. 2x + 3y = 18 There will be more presentations coming so keep checking your email.

More Related