Question 826787
Find an equation of the line that passes through the given point and is perpendicular to the given line. Write the equation in slope-intercept form.
(8, −4), 5x = 7y − 8
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First, find slope of:
5x = 7y − 8
5x+8 = 7y
(5/7)x+8/7 = y
slope is 5/7
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slope of "perpendicular" is negative reciprocal of 5/7:
slope of new line then is 
-7/5
.
using slope (-7/5) and (8,-4)
plug into "point-slope" form:
y - y1 = m(x - x1)
y - (-4) = (-7/5)(x - 8)
y + 4 = (-7/5)(x - 8)
y + 4 = (-7/5)x - (-7/5)8
y + 4 = (-7/5)x + 56/5
y = (-7/5)x + 56/5 - 4
y = (-7/5)x + 56/5 - 20/5
y = (-7/5)x + (56-20)/5
y = (-7/5)x + 36/5