## College Algebra (10th Edition)

$\color{blue}{y=-\dfrac{1}{2}x+\dfrac{5}{2}}$.
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 inverse reciprocal slopes. Line $L$ is perpendicular to the line $y=2x$, whose slope is $2$. Since $L$ is perpendicular to this line, $L$ has a slope of $-\dfrac{1}{2}$. Thus, the tentative equation of the line is $y=-\dfrac{1}{2}x+b$. To find the value of $b$, substitute the x and y values of the point $(1, 2)$ into the tentative equation above to obtain: $y=-\dfrac{1}{2}x+b \\2=-\dfrac{1}{2}(1) + b \\2=-\dfrac{1}{2}+ b \\2+\dfrac{1}{2}=b \\\dfrac{4}{2}+\dfrac{1}{2}=b \\\dfrac{5}{2}=b$ Therefore, the equation of line $L$ is $\color{blue}{y=-\dfrac{1}{2}x+\dfrac{5}{2}}$.