Answer
(a) Standard Form:
$x+3y=11$
(b) Slope-intercept form:
$y=-\frac{1}{3}x+\frac{11}{3}$
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 parallel to $x+3y=5$. Solve the equation for $y$ to obtain:
$3y=-x+5
\\\frac{3y}{3}=\frac{-x+5}{3}
\\y=-\frac{1}{3}x+\frac{5}{3}$
The slope of the line above is $-\frac{1}{3}$. Thus, the parallel of the line parallel to it is also $-\frac{1}{3}$.
This means that the tentative equation of the line we are looking for is $y=-\frac{1}{3}x+b$.
To find the value of $b$, substitute the $x$ and $y$ values of the point $(-1, 4)$ to obtain:
$y=-\frac{1}{3}x+b
\\4=-\frac{1}{3}(-1)+b
\\4=\frac{1}{3}+b
\\4-\frac{1}{3}=b
\\\frac{12}{3}-\frac{1}{3} =b
\\\frac{11}{3}=b$
Therefore, the equation of the line is $y=-\frac{1}{3}x+\frac{11}{3}$.
(a) Standard form
Multiply $3$ to both sides of the equation to obtain:
$3(y)=3(-\frac{1}{3}x+\frac{11}{3})
\\3y=-x+11$
Add $x$ to both sides of the equation to obtain:
$x+3y=-x+11+x
\\x+3y=11$
(b) Slope-intercept form
$y=-\frac{1}{3}x+\frac{11}{3}$
