Write an equation for a line passing through the point (4,7) that is parallel to y=(2)/(3)x+1 Then write a second equation 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 (4,7) that is parallel to y=(2)/(3)x+1 is square (Simplify your answer. Type your answer in slope -intercept form. Use integers or fractions for any numbers in the equation.) A second slope-intercept equation for a line passing through the point (4,7) that is perpendicular to the given line is square (Simplify your answer. Type your answer in slope -intercept form. Use integers or fractions for any numbers in the equation.)

To find the equation of a line passing through the point $(4,7)$ that is parallel to $y=\frac {2}{3}x+1$, we need to use the slope-intercept form of a linear equation, which is $y=mx+b$, where $m$ is the slope and $b$ is the y-intercept.<br /><br />Since the given line is $y=\frac {2}{3}x+1$, we know that the slope of this line is $\frac {2}{3}$. Since parallel lines have the same slope, the slope of the line we are looking for is also $\frac {2}{3}$.<br /><br />Now, we can substitute the slope and the given point into the slope-intercept form to solve for the y-intercept:<br /><br />$7 = \frac {2}{3}(4) + b$<br /><br />$7 = \frac {8}{3} + b$<br /><br />$b = 7 - \frac {8}{3}$<br /><br />$b = \frac {21}{3} - \frac {8}{3}$<br /><br />$b = \frac {13}{3}$<br /><br />So, the equation of the line passing through the point $(4,7)$ that is parallel to $y=\frac {2}{3}x+1$ is $y=\frac {2}{3}x+\frac {13}{3}$.<br /><br />To find the equation of a line passing through the point $(4,7)$ that is perpendicular to the given line, we need to find the negative reciprocal of the slope of the given line.<br /><br />The slope of the given line is $\frac {2}{3}$, so the negative reciprocal is $-\frac {3}{2}$.<br /><br />Now, we can substitute the slope and the given point into the slope-intercept form to solve for the y-intercept:<br /><br />$7 = -\frac {3}{2}(4) + b$<br /><br />$7 = -6 + b$<br /><br />$b = 7 + 6$<br /><br />$b = 13$<br /><br />So, the equation of the line passing through the point $(4,7)$ that is perpendicular to $y=\frac {2}{3}x+1$ is $y=-\frac {3}{2}x+13$.