Line l1 represents the graph of 3x + 4y = -14. Line l2 passes through the point (-5,7), and is perpendicular to line l1. If line l2 represents the graph of y=mx +b, then find m+b.
+171 Line l1 represents the graph of 3x + 4y = -14. Line l2 passes through the point (-5,7), and is perpendicular to line l1. If line l2 represents the graph of y=mx +b, then find m+b.
first of all lets find its perp line
$3x + 4y = -14 \implies 4x-3y=$perp line where given the condition that that perp line should go through $(-5,7)$, respectively $x=-5$ and $y=7$:
$4\left(-5\right)-3\left(7\right)=x$
$-20-21=x$
$-41=x$
thus, the perp line which goes through $(-5,7)$ is $4x-3y=-41$ BUT the thing is we want that in the form of $y=mx+b$
knowing what we have is in the form of $Ax + By = C$ to find $m$, the slope, we have to do $\frac{A}{B}$ so $\frac{4}{3} $
the y intercep, or $b$ is $\frac{-C}{B}$ so $\frac{-(-41)}{3} \implies\frac{41}{3} $
so the line in $y=mx+b$ form is $ y=\frac{4}{3}x+\frac{41}{3} $
Finally, lets find $m+b$:
$\frac{4}{3}+\frac{41}{3} =\boxed{\frac{45}{3}} \Rightarrow \boxed{15} $
here is an image of it as well:
