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.

Guest Jul 3, 2021

#1**+2 **

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:

UsernameTooShort Jul 3, 2021