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 $m+b$.
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
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 $m + b = \frac{4}{3} + \frac{41}{3} = \boxed{15}.$
Well, first off, let's find the slope of line $l_1$. We can do this by rearranging the equation $3x + 4y = -14$ to solve for $y$:
$4y = -3x - 14$
$y = -\frac{3}{4}x - \frac{14}{4}$
So the slope of line $l_1$ is $-\frac{3}{4}$. Since line $l_2$ is perpendicular to line $l_1$, its slope will be the negative reciprocal of $-\frac{3}{4}$, which is $\frac{4}{3}$.
We are given that line $l_2$ passes through the point $(-5,7)$. Using the point-slope form of a linear equation, we can write the equation for line $l_2$ as:
$y - 7 = \frac{4}{3}(x + 5)$
Simplifying this equation, we get:
$y = \frac{4}{3}x + \frac{20}{3}$
So $m+b = \frac{4}{3} + \frac{20}{3} = \frac{24}{3} = 8$.
Therefore, $m + b = 8$.
To find the equation of line $l_2$, let's first find the slope of line $l_1$.
The equation of line $l_1$ is given as $3x + 4y = -14$. We can rewrite this equation in slope-intercept form ($y = mx + b$) by isolating $y$:
$4y = -3x - 14$
$y = -\frac{3}{4}x - \frac{14}{4}$
Now we can see that the slope of line $l_1$ is $-\frac{3}{4}$.
Since line $l_2$ is perpendicular to line $l_1$, the slope of line $l_2$ is the negative reciprocal of $-\frac{3}{4}$.
The negative reciprocal of $-\frac{3}{4}$ is $\frac{4}{3}$.
Now we need to find the equation of line $l_2$ that passes through the point $(-5, 7)$ and has a slope of $\frac{4}{3}$.
Using the point-slope form of a line, the equation of line $l_2$ is:
$y - y_1 = m(x - x_1)$
where $(x_1, y_1)$ is the given point on line $l_2$ and $m$ is the slope.
Plugging in the values $(-5, 7)$ and $\frac{4}{3}$ for $x_1, y_1,$ and $m$ respectively, we get:
$y - 7 = \frac{4}{3}(x - (-5))$
$y - 7 = \frac{4}{3}(x + 5)$
$y - 7 = \frac{4}{3}x + \frac{4}{3} \cdot 5$
$y - 7 = \frac{4}{3}x + \frac{20}{3}$
Finally, let's rewrite this equation in slope-intercept form:
$y = \frac{4}{3}x + \frac{20}{3} +7$
$y = \frac{4}{3}x + \frac{20}{3} +\frac{21}{3}$
$y = \frac{4}{3}x + \frac{41}{3}$
From this equation, we can identify that the slope, $m$, is $\frac{4}{3}$, and the y-intercept, $b$, is $\frac{41}{3}$.
To find $m + b$, we add $\frac{4}{3}$ and $\frac{41}{3}$:
$m + b = \frac{4}{3} + \frac{41}{3} = \frac{45}{3} = \frac{15}{1} = 15$
Therefore, $m + b = 15$.
To find the equation of line $l_2$, we first need to determine the slope of line $l_1$. The equation of a line in the form $Ax + By = C$ can be rewritten as $y = -\frac{A}{B}x - \frac{C}{B}$, where the coefficient of $x$ is the slope of the line. In this case, we have $3x + 4y = -14$, so rewriting it in slope-intercept form gives $y = -\frac{3}{4}x - \frac{14}{4}$. Thus, the slope of line $l_1$ is $-\frac{3}{4}$.
Since line $l_2$ is perpendicular to line $l_1$, the slope of line $l_2$ will be the negative reciprocal of the slope of line $l_1$. Therefore, the slope of line $l_2$ is $\frac{4}{3}$.
Since line $l_2$ passes through the point $(-5,7)$, we can use the point-slope form of a line to find its equation. The point-slope form of a line is given by $y - y_1 = m(x - x_1)$, where $(x_1, y_1)$ is a point on the line and $m$ is the slope of the line. Plugging in the values $x_1 = -5$, $y_1 = 7$, and $m = \frac{4}{3}$, we get $y - 7 = \frac{4}{3}(x - (-5))$.
Simplifying the equation, we have $y - 7 = \frac{4}{3}(x + 5)$. Multiplying both sides by $3$ to eliminate the fractions, we get $3y - 21 = 4(x + 5)$.
Expanding and simplifying further, we have $3y - 21 = 4x + 20$. Rearranging the equation to the slope-intercept form, we get $y = \frac{4}{3}x + \frac{41}{3}$.
Therefore, the equation of line $l_2$ is $y = \frac{4}{3}x + \frac{41}{3}$, and $m + b = \frac{4}{3} + \frac{41}{3} = \frac{45}{3} = 15$.