SOLUTION: What is the equation of a line that is perpendicular to 7x+8y-12=0 and passes through (-2,5)?

Question 321157: What is the equation of a line that is perpendicular to 7x+8y-12=0 and passes through (-2,5)?
Found 2 solutions by Alan3354, mananth:
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
A line and a point example.
Write in standard form the eqation of the line that satisfies the given conditions. Perpendicular to 9x+3y=36, through (1,2)
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Find the slope of the line. Do that by putting the equation in slope-intercept form, y = mx + b. That means solve for y.
9x+3y = 36
3y= - 9x + 36
y = -3x + 12
The slope, m = -3
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The slope of lines parallel is the same.
The slope of lines perpendicular is the negative inverse, m = +1/3
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Use y = mx + b and the point (1,2) to find b.
2 = (1/3)*1 + b
b = 5/3
The equation is y = (1/3)x + 5/3 (slope-intercept form)
x - 3y = -5 (standard form)
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Answer by mananth(16946) (Show Source): You can put this solution on YOUR website!
7x+8y-12=0
8y=-7x+12
y=-7/8 x +12/8
..
slope = -7/8
the perpendicular to this line will have a slope of of 8/7
This line passes through (-2,5)
equation of line in slope intercept form =
y-y1= m(x-x1)
y-(-2)= 8/7(x-5)
y+2=8/7(x-5)
7(y+2)=8(x-5)
7y+14=8x-40
7y=8x-56
y=(8/7)x -8 is the equation of the required line
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