# How do you write the standard form of a line given (-6,7) and perpendicular to 3x+4y-12=0?

Jul 25, 2017

$4 x - 3 y = - 45$

#### Explanation:

Since the equation must be expressed in standard form, that is, $A x + B y = C$, we can find the perpendicular line by switching the coefficients of $x$ and $y$ and changing the operation sign.

But first let's rewrite the given equation in $A x + B y = C$ form.

$3 x + 4 y - 12 = 0$

$3 x + 4 y = 12$

To find the perpendicular line, switch the coefficients of $x$ and $y$ and change the operation sign. By doing so we get:

$4 x - 3 y = C$; where $C$ is any constant but we are given the point $\left(- 6 , 7\right)$ so we can substitute these values for $x$ and $y$ to find $C$.

If we let $\left(- 6 , 7\right) \to \left(x , y\right)$ then,

$4 \left(- 6\right) - 3 \left(7\right) = C$

$- 24 - 21 = C$

$- 45 = C$

So we found our constant. We can now substitute this value for $C$ for the equation of the perpendicular line. In doing so, we get our final equation:

$4 x - 3 y = - 45 \leftarrow$ And this is the equation of the perpendicular line