parallel+and+perpendicular+example

parallel+and+perpendicular+example - (9, -5); 5x + 3y = 4...

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Write an equation of the line containing the given point and parallel to the given line. (9, -5); 5x + 3y = 4 First, you must put the equation into y = mx + b form. 5x + 3y = 4 -5x -5x 3y = -5x + 4 Now you must divide each term by 3. 3 3 3 Y = - 53 x + 43 The slope of the given line is - 53 . In order to find an equation that will be parallel, you must use this slope. We will use that for our slope now with the given point, (9, -5), in the following formula to find the equation of the parallel line: y – y1 = m(x –x1) y – (-5) = - 53 (x – 9) y + 5 = - 53 x + 453 Reduce 453 before subtracting 5 from both sides. y + 5 = - 53 x + 15 -5 -5 Y = - 53 x + 10 This is the equation of the line perpendicular to 5x + 3y = 4.
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Write an equation of the line containing the given point and perpendicular to the given line.
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Unformatted text preview: (9, -5); 5x + 3y = 4 First, you must put the equation into y = mx + b form. 5x + 3y = 4-5x -5x 3y = -5x + 4 Now you must divide each term by 3. 3 3 3 Y = -53x + 43 The slope of the given line is - 53. In order to find an equation that will be perpendicular, you must use the negative reciprocal of this slope. This would be – (- ⅗) or +⅗. We will use that for our slope now with the given point, (9, -5), in the following formula to find the equation of the perpendicular line: y – y1 = m(x –x1) y – (-5) = ⅗(x – 9) y + 5 = ⅗x - 275-5 -5 We must change 5 here to a fraction with a common denominator before subtracting. This would be 255. Y = ⅗x - 275 - 255 Y = ⅗x - 525 This is the equation of the line perpendicular to 5x + 3y = 4....
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This note was uploaded on 08/12/2010 for the course MAT 116 taught by Professor Universityofphoenix during the Spring '09 term at University of Phoenix.

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parallel+and+perpendicular+example - (9, -5); 5x + 3y = 4...

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