Write an equation for a line passing through the point (7,5) that is parallel to y=(2)/(3)x+6 Then write a sec for a line passing through the given point that is perpendicular to the given line A slope-intercept equation for a line passing through the point (7,5) that is parallel to y=(2)/(3)x+6 is square (Simplify your answer. Type your answer in slope intercept form. Use integers or fractions for any numbers the equation.)

## Step 1
The given line is $y=\frac{2}{3}x+6$. The slope of this line is $\frac{2}{3}$.

## Step 2
A line parallel to another line has the same slope. Therefore, the slope of the line we are looking for is also $\frac{2}{3}$.

## Step 3
We know that the line we are looking for passes through the point (7,5). We can use the slope-intercept form of a line, which is $y = mx + b$, where $m$ is the slope and $b$ is the y-intercept.

## Step 4
Substitute the slope $\frac{2}{3}$ and the point (7,5) into the equation $y = mx + b$ to solve for $b$.

### $5 = \frac{2}{3} * 7 + b$

## Step 5
Solve the equation for $b$.

### $b = 5 - \frac{14}{3} = \frac{15}{3} - \frac{14}{3} = \frac{1}{3}$

## Step 6
Substitute the slope $\frac{2}{3}$ and the y-intercept $\frac{1}{3}$ into the slope-intercept form of a line to get the equation of the line.

### $y = \frac{2}{3}x + \frac{1}{3}$