## Calculus 10th Edition

$a)$ $2x-y-3=0$ $b)$ $x+2y-4=0$
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: $\dfrac{1}{2}x+y-2=0$ $2\Big(\dfrac{1}{2}x+y-2=0\Big)$ $x+2y-4=0$