math.answers.com/math-and-arithmetic/How_can_you_use_algebra_for_perpendicular_lines
Preview meta tags from the math.answers.com website.
Linked Hostnames
8- 33 links tomath.answers.com
- 19 links towww.answers.com
- 1 link totwitter.com
- 1 link towww.facebook.com
- 1 link towww.instagram.com
- 1 link towww.pinterest.com
- 1 link towww.tiktok.com
- 1 link towww.youtube.com
Thumbnail

Search Engine Appearance
How can you use algebra for perpendicular lines? - Answers
Using algebra to determine if two lines are perpendicular to one another we first must determine each line's slope. Select two known points on each line to determine the slope for the line. The Point-slope form of a linear equation is (Y1-Y2) = m(X1-X2). Therefore The slope m = (Y1-Y2)/(X1-X2) We will use these points to generate the slope equation. Line A Line B Point 1 Point 2 Point 1 Point 2 X1,Y1 X2,Y2 A1,B1 A2,B2 If the product of the slopes of two lines = -1 then the two lines are perpendicular. Using the point slope form above the equation would look like this: [(Y1-Y2)/(X1-X2)] X [(A1-A2)/(B1-B2)] = m(line A) X m(line B) Example Line A Line B Point 1 Point 2 Point 1 Point 2 0,0 3,3 3,-3 0,0 Using the above formula [(0-3)/(0-3)] X [(3-0)/(-3-0)] = [-3/-3] X [3/-3] = 1 x -1 = -1 These two lines are perpendicular.
Bing
How can you use algebra for perpendicular lines? - Answers
Using algebra to determine if two lines are perpendicular to one another we first must determine each line's slope. Select two known points on each line to determine the slope for the line. The Point-slope form of a linear equation is (Y1-Y2) = m(X1-X2). Therefore The slope m = (Y1-Y2)/(X1-X2) We will use these points to generate the slope equation. Line A Line B Point 1 Point 2 Point 1 Point 2 X1,Y1 X2,Y2 A1,B1 A2,B2 If the product of the slopes of two lines = -1 then the two lines are perpendicular. Using the point slope form above the equation would look like this: [(Y1-Y2)/(X1-X2)] X [(A1-A2)/(B1-B2)] = m(line A) X m(line B) Example Line A Line B Point 1 Point 2 Point 1 Point 2 0,0 3,3 3,-3 0,0 Using the above formula [(0-3)/(0-3)] X [(3-0)/(-3-0)] = [-3/-3] X [3/-3] = 1 x -1 = -1 These two lines are perpendicular.
DuckDuckGo
How can you use algebra for perpendicular lines? - Answers
Using algebra to determine if two lines are perpendicular to one another we first must determine each line's slope. Select two known points on each line to determine the slope for the line. The Point-slope form of a linear equation is (Y1-Y2) = m(X1-X2). Therefore The slope m = (Y1-Y2)/(X1-X2) We will use these points to generate the slope equation. Line A Line B Point 1 Point 2 Point 1 Point 2 X1,Y1 X2,Y2 A1,B1 A2,B2 If the product of the slopes of two lines = -1 then the two lines are perpendicular. Using the point slope form above the equation would look like this: [(Y1-Y2)/(X1-X2)] X [(A1-A2)/(B1-B2)] = m(line A) X m(line B) Example Line A Line B Point 1 Point 2 Point 1 Point 2 0,0 3,3 3,-3 0,0 Using the above formula [(0-3)/(0-3)] X [(3-0)/(-3-0)] = [-3/-3] X [3/-3] = 1 x -1 = -1 These two lines are perpendicular.
General Meta Tags
22- titleHow can you use algebra for perpendicular lines? - Answers
- charsetutf-8
- Content-Typetext/html; charset=utf-8
- viewportminimum-scale=1, initial-scale=1, width=device-width, shrink-to-fit=no
- X-UA-CompatibleIE=edge,chrome=1
Open Graph Meta Tags
7- og:imagehttps://st.answers.com/html_test_assets/Answers_Blue.jpeg
- og:image:width900
- og:image:height900
- og:site_nameAnswers
- og:descriptionUsing algebra to determine if two lines are perpendicular to one another we first must determine each line's slope. Select two known points on each line to determine the slope for the line. The Point-slope form of a linear equation is (Y1-Y2) = m(X1-X2). Therefore The slope m = (Y1-Y2)/(X1-X2) We will use these points to generate the slope equation. Line A Line B Point 1 Point 2 Point 1 Point 2 X1,Y1 X2,Y2 A1,B1 A2,B2 If the product of the slopes of two lines = -1 then the two lines are perpendicular. Using the point slope form above the equation would look like this: [(Y1-Y2)/(X1-X2)] X [(A1-A2)/(B1-B2)] = m(line A) X m(line B) Example Line A Line B Point 1 Point 2 Point 1 Point 2 0,0 3,3 3,-3 0,0 Using the above formula [(0-3)/(0-3)] X [(3-0)/(-3-0)] = [-3/-3] X [3/-3] = 1 x -1 = -1 These two lines are perpendicular.
Twitter Meta Tags
1- twitter:cardsummary_large_image
Link Tags
16- alternatehttps://www.answers.com/feed.rss
- apple-touch-icon/icons/180x180.png
- canonicalhttps://math.answers.com/math-and-arithmetic/How_can_you_use_algebra_for_perpendicular_lines
- icon/favicon.svg
- icon/icons/16x16.png
Links
58- https://math.answers.com
- https://math.answers.com/math-and-arithmetic/How_can_you_use_algebra_for_perpendicular_lines
- https://math.answers.com/math-and-arithmetic/How_do_you_find_the_height_of_a_rectangular_prism_given_surface_area_length_and_width
- https://math.answers.com/math-and-arithmetic/How_many_times_will_26_go_into_111
- https://math.answers.com/math-and-arithmetic/Is_21._21