Question

Let line $l_1$ be the graph of $3x + 4y = -14$. Line $l_2$ is perpendicular to line $l_1$ and passes through the point $(-5,7)$. If line $l_2$ is the graph of the equation $y=mx +b$, then find $m+b$.

Answer 1

$3x+4y=-14\implies y=-\dfrac{3x+14}4$

and so $\ell_1$ has slope -3/4. Then any line perpendicular to $\ell_1$ has slope 4/3.

Given that $\ell_2\perp\ell_1$ and $\ell_2$ passes through (-5, 7), its equation is

$y-7=\dfrac43(x+5)\implies y=\dfrac{4x}3+\dfrac{41}3$

so that $m=\frac43$ and $b=\frac{41}3$, which gives

$m+b=\dfrac{45}3=\boxed{15}$