Answer
a) $y=-\dfrac{5}{2}x+2$
b) $y=\dfrac{2}{5}x+\dfrac{13}{5}$
Work Step by Step
a) In the form $y=mx+b$, the given equation $
5x+2y=6
,$ is equivalent to
\begin{align*}\require{cancel}
2y&=-5x+6
\\\\
\dfrac{\cancel2y}{\cancel2}&=\dfrac{-5x+6}{2}
\\\\
y&=\dfrac{-5}{2}x+\dfrac{6}{2}
\\\\
y&=-\dfrac{5}{2}x+3
.\end{align*}
Since the slope of $y=mx+b$ is given by $m$, then the slope of the line in the equation above is $-\dfrac{5}{2}$.
Since parallel lines have the same slope, then the slope of the required line is $m=-\dfrac{5}{2}$. Using $y-y_1=m(x-x_1)$ or the Point-Slope Form of linear equations, with $(x_1,y_1)=(2,-3),$ then
\begin{align*}
y-(-3)&=-\dfrac{5}{2}(x-2)
\\\\
y+3&=-\dfrac{5}{2}(x-2)
.\end{align*}
In the form $y=mx+b,$ the equation above is equivalent to
\begin{align*}
y+3&=-\dfrac{5}{2}x+5
\\\\
y&=-\dfrac{5}{2}x+5-3
\\\\
y&=-\dfrac{5}{2}x+2
.\end{align*}
Hence, the equation of the required line is $
y=-\dfrac{5}{2}x+2
$.
b) In the form $y=mx+b$, the given equation $
5x+2y=6
,$ is equivalent to
\begin{align*}\require{cancel}
2y&=-5x+6
\\\\
\dfrac{\cancel2y}{\cancel2}&=\dfrac{-5x+6}{2}
\\\\
y&=\dfrac{-5}{2}x+\dfrac{6}{2}
\\\\
y&=-\dfrac{5}{2}x+3
.\end{align*}
Since the slope of $y=mx+b$ is given by $m$, then the slope of the line in the equation above is $-\dfrac{5}{2}$.
Since perpendicular lines have negative reciprocal slopes, then the slope of the required line is $m=\dfrac{2}{5}$. Using $y-y_1=m(x-x_1)$ or the Point-Slope Form of linear equations, with $(x_1,y_1)=(-4,1),$ then
\begin{align*}
y-1&=\dfrac{2}{5}(x-(-4))
\\\\
y-1&=\dfrac{2}{5}(x+4)
.\end{align*}
In the form $y=mx+b,$ the equation above is equivalent to
\begin{align*}
y-1&=\dfrac{2}{5}(x)+\dfrac{2}{5}(4)
\\\\
y-1&=\dfrac{2}{5}x+\dfrac{8}{5}
\\\\
y&=\dfrac{2}{5}x+\dfrac{8}{5}+1
\\\\
y&=\dfrac{2}{5}x+\dfrac{8}{5}+\dfrac{5}{5}
\\\\
y&=\dfrac{2}{5}x+\dfrac{13}{5}
.\end{align*}
Hence, the equation of the required line is $
y=\dfrac{2}{5}x+\dfrac{13}{5}
$.
