## College Algebra (10th Edition)

$y=-\frac{1}{2}x+\frac{13}{2}$
RECALL: (1) The slope-intercept form of a line's equation is: $y=mx+b$ where $m$ = slope and $b$ = y-intercept (2) Perpendicular lines have slopes whose product is $-1$. The line we are looking for is perpendicular to the given line $y=2x+1$, whose slope is $2$. This means that the slope of the line we are looking for has a slope of $-\frac{1}{2}$ since $2(-\frac{1}{2}) = -1$. Thus, the tentative equation of the line is: $y =-\frac{1}{2}x+b$ To find the value of $b$, substitute the x and y values of the point $(3, 5)$ to obtain: $y=-\frac{1}{2}x+b \\5 = -\frac{1}{2}(3) + b \\5 = -\frac{3}{2} + b \\5+\frac{3}{2}=b \\\frac{10}{2} + \frac{3}{2} = b \\\frac{13}{2}=b$ Therefore, the equation of the line perpendicular to the given line is: $y=-\frac{1}{2}x+\frac{13}{2}$