Let line \(l_1\) be the graph of \(5x + 8y = -9 \) . Line \(l_2\)is perpendicular to line \(1_1\)and passes through the point \((10,10)\). If line \(1_2\) is the graph of the equation \(y = mx +b\) , then find \(m+b\).
L1 = 5x + 8y = -9
In the form Ax + By = C....the slope of this line is -A / B
So...the slope of this line = -5/8
And the slope of a line perpendicular to this = 8/5 = slope of L2
So....we have this equation of L2 that passes through (10, 10)
y = (8/5)(x -10) + 10
y = (8/5)x - 16 + 10
y = (8/5)x - 6
m = 8/5 b = -6 and their sum is 8/5 + (-6) = 8/5 + (-30/5) = -22/5