# How do you write the standard form of the equation of the line that is perpendicular to y=-4/3x-7 and contains (-5,-7)?

Jun 27, 2018

The standard form equation is $3 x - 4 y = 13$.

#### Explanation:

The slope of our line is $\textcolor{red}{- \setminus \frac{4}{3}}$. The slope of any perpendicular line is given by the negative reciprocal.

$m ' = - \setminus \frac{1}{m}$

$m ' = \setminus \frac{1}{\textcolor{red}{- \setminus \frac{4}{3}}}$

$m ' = \setminus \frac{3}{4}$

The second condition is that the line contains $\left(\textcolor{b l u e}{- 5} , \textcolor{g r e e n}{- 7}\right)$.

$y = \frac{3}{4} x + b$

Substitute $\textcolor{b l u e}{x = - 5}$, $\textcolor{g r e e n}{y = - 7}$.

$\textcolor{g r e e n}{- 7} = \frac{3}{4} \left(\textcolor{b l u e}{- 5}\right) + b$

$\textcolor{g r e e n}{- 7} = - \frac{15}{4} + b$

$- \frac{13}{4} = b$

So in slope-intercept form, the equation of our line is

$y = \frac{3}{4} x - \frac{13}{4}$

but the question requires it in standard form. This means that it should be in the form

$A x + B y = C$

where $A$, $B$, and $C$ are all integers and $A$ is positive. In our equation, we will multiply by the lowest common denominator to eliminate the denominator.

$4 y = 3 x - 13$

$\iff 3 x - 13 = 4 y$

$\iff 3 x - 4 y = 13$