#### Answer

$a)$ $2x-y-3=0$
$b)$ $x+2y-4=0$

#### Work Step by Step

Point $(2,1)$ $;$ Line $4x-2y=3$
$a)$ Find a line through the point given and parallel to the line given
Solve the equation of the line given for $y$:
$4x-2y=3$
$-2y=-4x+3$
$y=\dfrac{-4}{-2}x-\dfrac{3}{2}$
$y=2x-\dfrac{3}{2}$
The line given is now in slope-intercept form, which is $y=mx+b$, where $m$ is the slope of the line and $b$ is its $y$-intercept. From this equation, it can be seen that the slope of the line given is $m=2$
Since parallel lines have the same slope, the slope of the line to be found is also $m=2$
The slope of the line to be found and a point through which it passes are now known. Use the point-slope form of the equation of a line, which is $y-y_{1}=m(x-x_{1})$, where $m$ is the slope of the line and $(x_{1},y_{1})$ is a point through which it passes, to obtain the equation. Substitute the known values into the formula and simplify:
$y-1=2(x-2)$
$y-1=2x-4$
$y=2x-4+1$
$y=2x-3$
Since the equation must be given in general form, take all terms to the right side and rearrange to obtain the answer:
$0=2x-y-3$
$2x-y-3=0$
$b)$ Find a line through the point given and perpendicular to the line given
The slope of the line given was found in part $a$, and it is $m=2$.
The slopes of perpendicular lines are negative reciprocals, so the slope of the line to be found is $m=-\dfrac{1}{2}$
The slope of the line to be found and a point through which it passes are now known. Use the point-slope form of the equation of a line, which is $y-y_{1}=m(x-x_{1})$, where $m$ is the slope of the line and $(x_{1},y_{1})$ is a point through which it passes, to obtain the equation. Substitute the known values into the formula and simplify:
$y-1=-\dfrac{1}{2}(x-2)$
$y-1=-\dfrac{1}{2}x+1$
$y=-\dfrac{1}{2}x+1+1$
$y=-\dfrac{1}{2}x+2$
Since the equation must be given in general form, take all terms to the left side and multiply the whole equation by $2$ to obtain the answer:
$\dfrac{1}{2}x+y-2=0$
$2\Big(\dfrac{1}{2}x+y-2=0\Big)$
$x+2y-4=0$