Question 7(Multiple Choice Worth 1 points) (02.04 MC) Write the equation of the line that passes through (3,-2) and is perpendicular to y=(3)/(4)x+6 y=-(4)/(3)x+(1)/(3) y=-(4)/(3)x+2 y=(3)/(4)x-(17)/(4) y=(4)/(3)x-6

## Step 1
The is $y=\frac{3}{4}x+6$. The slope of this line is $\frac{3}{4}$.

## Step 2
The slope of a line perpendicular to a given line is the negative reciprocal of the slope of the given line. Therefore, the slope of the line perpendicular to the given line is $-\frac{4}{3}$.

## Step 3
The equation of a line in slope-intercept form is $y=mx+b$, where $m$ is the slope and $b$ is the y-intercept.

## Step 4
We know that the line passes through the point (3,-2). Substituting these values into the equation, we get $-2=-\frac{4}{3}*3+b$.

## Step 5
Solving for $b$, we get $b=2$.

## Step 6
Therefore, the equation of the line that passes through (3,-2) and is perpendicular to $y=\frac{3}{4}x+6$ is $y=-\frac{4}{3}x+2$.