Answer
$y=-2x+\dfrac{1}{3}$, or, in general form, $6x+3y-1=0$
Work Step by Step
Through $(\frac{1}{2},-\frac{2}{3});$ perpendicular to the line $4x-8y=1$
A point through which the line passes and the fact that the line is perpendicular to another line whose equation is $4x-8y=1$ are known.
Use the point-slope form of the equation of a line, which is $y-y_{1}=m(x-x_{1})$, where $(x_{1},y_{1})$ is a point through which the line passes and $m$ is the slope.
Solve the equation of the given line for $y$. The resulting expression will be in the slope-intercept form and the slope can be identified as the number in front of $x$:
$4x-8y=1$
$-8y=-4x+1$
$y=\dfrac{-4}{-8}x-\dfrac{1}{8}$
$y=\dfrac{1}{2}x-\dfrac{1}{8}$
The slope of the given line is $\dfrac{1}{2}$
Since the slopes of perpendicular lines are negative reciprocals, the slope of the line whose equation must be found is:
$m=-\dfrac{1}{\Big(\dfrac{1}{2}\Big)}=-2$
Substitute $(x_{1},y_{1})$ and $m$ into the point-slope form of the equation of a line formula and simplify to obtain the answer:
$y-y_{1}=m(x-x_{1})$
$y+\dfrac{2}{3}=-2\Big(x-\dfrac{1}{2}\Big)$
$y+\dfrac{2}{3}=-2x+1$
$y=-2x+1-\dfrac{2}{3}$
$y=-2x+\dfrac{1}{3}$, or, in general form, $6x+3y-1=0$
