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 $m+b$.

Answer 1

Let me rephrase your question without all that $ stuff

Line 1 represents the graph of 3x + 4y = -14. Line 2 passes through the point (-5,7), and is perpendicular to line 1. If line 2 represents the graph of y=mx +b, then find m+b.

The slope of line 1 is -3/4. So the slope of line 2 must be 4/3 and the equation is
y = (4/3)x + b
but (-5,7) is supposed to lie on it, then
7 = (4/3)(-5) + b
b = 41/3

then m+b = 4/3 + 41/3 = 15

Answer 2

Since lines $l_1$ and $l_2$ are perpendicular (and neither is vertical), the product of their slopes is $-1$. To find the slope of $l_1$, we write $3x + 4y = -14$ in slope-intercept form by solving for $y$. Solving for $y,$ we find
\[y = -\frac{3}{4} x - \frac{7}{2}.\]Therefore, the slope of $l_1$ is $-\frac{3}{4}$.

Since line $l_2$ is perpendicular to line $l_1$, the slope of $l_2$ is $\frac{4}{3}.$ The line $l_2$ passes through the point $(-5,7)$, so its equation is given by
\[y - 7= \frac{4}{3} (x + 5).\]Isolating $y,$ we find
\[y = \frac{4}{3} x + \frac{41}{3},\]so $m + b = \frac{4}{3} + \frac{41}{3} = \boxed{15}.$