## Precalculus (6th Edition)

(a) Standard form $5x-3y=-13$ (b) Slope-intercept form $y=\frac{5}{3}x+\frac{13}{3}$
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 $3x+5y=1$. Solve the equation for $y$ to obtain: $3x+5y=1 \\5y=1-3x \\5y=-3x+1 \\\frac{5y}{5}=\frac{-3x+1}{5} \\y=-\frac{3}{5}x+\frac{1}{5}$ The slope of the line above is $-\frac{3}{5}$. Thus, the slope of the line perpendicular to it is $\frac{5}{3}$. This means that the tentative equation of the line we are looking for is $y=\frac{5}{3}x+b$. To find the value of $b$, substitute the $x$ and $y$ values of the point $(1, 6)$ to obtain: $y=\frac{5}{3}x+b \\6=\frac{5}{3}(1)+b \\6=\frac{5}{3}+b \\6-\frac{5}{3}=b \\\frac{18}{3}-\frac{5}{3}=b \\\frac{13}{3}=b$ Therefore, the equation of the line is $y=\frac{5}{3}x+\frac{13}{3}$. (a) Standard form Multiply $3$ to both sides of the equation to obtain: $3(y)=3(\frac{5}{3}x+\frac{13}{3}) \\3y=5x+13$ Subtract $3y$ to both sides of the equation to obtain: $3y-3y=5x+13-3y \\0=5x-3y+13$ Subtract $13$ to both sides to obtain: $0-13=5x-3y+13-13 \\-13=5x-3y \\5x-3y=-13$ (b) Slope-intercept form $y=\frac{5}{3}x+\frac{13}{3}$