Let line \(l_1\) be the graph of \(5x + 8y = -9\). Line \(l_2\) is perpendicular to line \(l_1\) and passes through the point (10,10). If line \(l_2\) is the graph of the equation y=mx+b, then find m+b.
+128408 L1 = 5x + 8y = - 9
In the form Ax + By = C......the slope of this line = -A/B
So....the slope of L1 = -5 /8
So....the slope of perpendicular L2 = 8/5
So..... using the point (10,10)....the equation of L2 is
y = (8/5) (x - 10) + 10 simplify
y = (8/5)x - 16 + 10
y = (8/5)x - 6
m = 8/5 and b = -6
So....m + b = 8/5 - 6 = 8/5 - 30/5 = -22/5
