SOLUTION: What is the slope of any line perpendicular to the line formed by the equation -7x + 63y = 9?
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Question 432251: What is the slope of any line perpendicular to the line formed by the equation -7x + 63y = 9?
Found 2 solutions by algebrahouse.com, Alan3354:
Answer by algebrahouse.com(1659) (Show Source): You can put this solution on YOUR website!
"What is the slope of any line perpendicular to the line formed by the equation
-7x + 63y = 9?"
Slope-intercept form is y = mx + b
m is the slope
b is the y-intercept
-7x + 63y = 9
63y = 7x + 9 {added 7x to both sides}
y = (1/9)x + 1/7 {divided both sides by 63}
slope = 1/9
Perpendicular lines have slopes which are negative reciprocals.
If the slope of this line is 1/9,
the perpendicular line would have a slope of -9.
-9 is the answer
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Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
-7x + 63y = 9
Solve for y
63y = 7x + 9
y = (1/9)x + 1/7
Slope m = 1/9
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The slope of lines perpendicular is the negative reciprocal = -9
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