## Precalculus (6th Edition)

(a) Standard form $3x+8y=-6$ (b) Slope-intercept form $y=-\frac{3}{8}x-\frac{3}{4}$
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 perpendicular to $8x-3y=7$. Solve the equation for $y$ to obtain: $8x-3y=7 \\8x-7=3y \\\frac{8x-7}{3}=\frac{3y}{3} \\\frac{8}{3}x-\frac{7}{3}=y \\y=\frac{8}{3}x-\frac{7}{3}$ The slope of the line above is $\frac{8}{3}$. Thus, the slope of the line perpendicular to it is $-\frac{3}{8}$. This means that the tentative equation of the line we are looking for is $y=-\frac{3}{8}x+b$. To find the value of $b$, substitute the $x$ and $y$ values of the point $(-2, 0)$ to obtain: $y=-\frac{3}{8}x+b \\0=-\frac{3}{8}(-2)+b \\0=\frac{3}{4}+b \\0-\frac{3}{4}=b \\-\frac{3}{4}=b$ Therefore, the equation of the line is $y=-\frac{3}{8}x-\frac{3}{4}$. (a) Standard form Multiply $8$ to both sides of the equation to obtain: $8(y)=8(-\frac{3}{8}x-\frac{3}{4}) \\8y=-3x-6$ Add $3x$ to both sides of the equation to obtain: $8y+3x=-3x-6+3x \\3x+8y=-6$ (b) Slope-intercept form $y=-\frac{3}{8}x-\frac{3}{4}$