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High School Mathematics9.15 Problems on Slope

 Example - I: The slope of a line passing through (5,3) and (6,a) is 8/3. Find the value of a. Solution: Given that, The line is passing through (5,3) and (6,a) Therefore, the slope of these two points is m = (y2 - y1) / (x2 - x1) m = (a - 3) / (6 - 5) m = a - 3 Given that, m = 8/3 a - 3 = 8/3 a = 8/3 - 3 a = (8 - 9)/3 a = -1/3 Example - II: The line 4x + 3y - 5 = 0 is perpendicular to 3x - ay + 7 = 0, find the value of a. Solution: Let the slope of 4x + 3y - 5 = 0, be m1, Then m1 = -x coefficient / y coefficient. i.e., m1 = -4/3 If m2 is the slope of 3x - ay + 7 = 0, then m2 = -3 / -a m2 = 3/a Given that the lines are perpendicular, we know that the product of the perpendicular lines slope is equal to -1. i.e., m1m2 = -1 -4/3 * 3/a = -1 -4/a = -1 a = 4 Direction: Solve the following problems. Also write at least 10 examples of your own.

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High School Mathematics9.15 Problems on Slope

 Q 1: The slope of a line passing through (4,5) and (7,a) is 4. Find the value of a.15171413 Q 2: The slope of a line passing through (3,4) and (a,7) is 15. Find the value of a.516/55/1616 Q 3: The slope of a line passing through (3,8) and (5,a) is 7/2. Find the value of a.12131514 Q 4: The slope of a line passing through (4,6) and (8,a) is 6. Find the value of a.30201525 Q 5: The line 4x + 7y - 5 = 0 is perpendicular to 3x - ay - 7 = 0, find the value of a.1277/1212/7 Q 6: The line 3x + 5y - 15 = 0 is perpendicular to ax + 3y - 5 = 0, find the value of a.5-53-3 Question 7: This question is available to subscribers only! Question 8: This question is available to subscribers only!

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