Question 986526
Let line l1 be the graph of 5x + 8y = -9. Line l2 is perpendicular to line l1 and passes through the point (10,10). If line l2 is the graph of the equation y=mx +b, then find m+b.
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Find the slope of L1 5x + 8y = -9
To do that, put it in slope-intercept form y = mx + b.  That means solve for y.
5x + 8y = -9
8y = -5x - 9
y = (-5/8)x - 9/8
m is the slope = -5/8
b is the y-intercept = -9/8 but that's not relevant here.
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The slope of any line perpendicular is the negative inverse of -5/8 --> 8/5.
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Now find the equation of L2 thru (10,10) with a slope of 8/5
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y-y1 = m*(x-x1) where (x1,y1) is the point (10,10)'
y-10 = (8/5)*(x-10) = (8/5)x - 80/5
y = (8/5)x - 6
--> for L2, m = 8/5 and b = -6