## Precalculus (6th Edition)

(a) Standard form $2x-y=8$ (b) Slope-intercept form $y=2x-5$
RECALL: (1) The standard form of a line's equation is $Ax+By=C$. (2) THe slope-intercept form of a line's equation is $y=mx+b$ where $m$=slope and $(0, b)$ is the line's y-intercept. (3) Parallel lines have the same slope. (4) Perpendicular lines have slopes whose product is $-1$ (negative reciprocals of each other). The line is parallel to $2x-y=5$. Solve the equation for $y$ to obtain: $2x-y+y=5+y \\2x=5+y \\2x-5=5+y-5 \\2x-5=y \\y=2x-5$ The slope of the line above is $2$. Thus, the slope of the line parallel to it is also $2$. This means that the tentative equation of the line we are looking for is $y=2x+b$. To find the value of $b$, substitute the $x$ and $y$ values of the point $(3, -2)$ to obtain: $y=2x+b \\-2=2(3)+b \\-2=6+b \\-2-6=b \\-8=b$ Therefore, the equation of the line is $y=2x-8$. (a) Standard form Add $8$ to both sides of the equation to obtain: $y+8=2x-8+8 y+8=2x$ Subtract $y$ to both sides of the equation to obtain: $y+8-y=2x-y \\8=2x-y \\2x-y=8$ (b) Slope-intercept form $y=2x-5$