Answer
(a) Standard form
$3x+8y=-6$
(b) Slope-intercept form
$y=-\frac{3}{8}x-\frac{3}{4}$
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 $8x-3y=7$. 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
$y=-\frac{3}{8}x-\frac{3}{4}$