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=-\frac{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=\frac43(x+5)\implies y=\frac{4x}3+\frac{41}3`

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

`m+b=\frac{45}3=\boxed{15}`