+47 Line $l_1$ represents the graph of $3x + 4y = -14$. Line $l_2$ passes through the point $(-5,7)$, and is perpendicular to line $l_1$. If line $l_2$ represents the graph of $y=mx +b$, then find $m+b$.
Isolate y in 3x + 4y = -14 ==> y = 3x/4 - 14/4
Slope of Perpendicular = -4/3, point-slope form of equation y - 7 = -4/3(x + 5)
==> 3y - 21 = -4(x + 5) = -4x + 20
==> 3y = -4x + 41
==> y = -4/3*x + 41/3
m + b = -4/3 + 41/3 = 37/3
+36915 L1 3x+4y = - 14 is tha same as y = - 3/4 x - 14/4 slope, m = - 3/4 perpindicular slope is - 1/m = 4/3
Your new line L2 is thus y = 4/3 x + b sub in the point given to calculate 'b'
7 = 4/3 (-5) + b
so b = 13 2/3
y = 4/3 x + 13 2/3 Guest messed up with the slope ...it is not negative in the perpindicular line....
you can finish...... 
