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https://math.answers.com/math-and-arithmetic/How_can_you_use_algebra_for_perpendicular_lines

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.



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How can you use algebra for perpendicular lines? - Answers

https://math.answers.com/math-and-arithmetic/How_can_you_use_algebra_for_perpendicular_lines

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.



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https://math.answers.com/math-and-arithmetic/How_can_you_use_algebra_for_perpendicular_lines

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.

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      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.
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