## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

$y-\frac{1}{2}=\frac{2}{3}(x)$
RECALL: (1) The point-slope form of a line's equation is $y-y_1=m(x-x_1)$ where $m$=slope and $(x_1, y_1)$ is a point on the line. (2) The slope-intercept form of a line's equation is $y=mx+b$ where $m$=slope and $b$ is the y-coordinate of the line's y-intercept. (3) Parallel lines have equal slopes. (4) Perpendicular lines have slopes whose product is $-1$ (negative reciprocals of each other). Write $2x-3y=4$ in slope-intercept form: $2x-3y=4 \\-3y=4-2x \\-3y=-2x-4 \\\frac{-3y}{-3}=\frac{-2x-4}{-3} \\y=\frac{2}{3}x+\frac{4}{3}$ This means the equation $2x-3y=4$ is equivalent to $y=\frac{2}{3}x+\frac{4}{3}$. The line is parallel to $y=\frac{2}{3}x+\frac{4}{3}$. Since the slope of this line is $\frac{2}{3}$, then the slope of the line parallel to it is also $\frac{2}{3}$. Using the given point on the line $(0, \frac{1}{2})$ and the slope $\frac{2}{3}$, the equation of the line in point-slope form is: $y-\frac{1}{2}=\frac{2}{3}(x-0) \\y-\frac{1}{2}=\frac{2}{3}(x)$