#### Answer

(a) Standard form
$5x-3y=-13$
(b) Slope-intercept form
$y=\frac{5}{3}x+\frac{13}{3}$

#### Work Step by Step

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
$y=\frac{5}{3}x+\frac{13}{3}$