## College Algebra (10th Edition)

$\color{blue}{y=\frac{1}{2}x}$
RECALL: (1) The slope-intercept form of a line's equation is: $y=mx+b$ where $m$ = slope and $b$ = y-intercept (2) Parallel lines have the same (equal) slopes. (3) Perpendicular lines have slopes whose product is $-1$ (negative reciprocals of each other). Write the given line in slope-intercept form to obtain: $x-2y=-5 \\x+5=2y \\\frac{x+5}{2} = \frac{2y}{2} \\\frac{x}{2} + \frac{5}{2}=y \\y=\frac{1}{2}x+\frac{5}{2}$ This line has a slope of $\frac{1}{2}$. Since the line we are looking for is parallel to the line above, then its slope is also $\\$ $\frac{1}{2}$. Thus, the tentative equation of the line is: $y=\frac{1}{2}x+b$ The line contains the point $(0, 0)$. This means that the y-intercept of the line is $0$. Therefore, the equation of the line is: $y=\frac{1}{2}x+0 \\\color{blue}{y=\frac{1}{2}x}$