#### Answer

$\text{slope-intercept form: }
y=-\dfrac{2}{5}x+\dfrac{13}{5}
\\\\
\text{standard form: }
2x+5y=13$

#### Work Step by Step

In the slope-intercept form, the given equation, $
2x+5y=10
,$ is equivalent to
\begin{array}{l}\require{cancel}
5y=-2x+10
\\\\
y=-\dfrac{2}{5}x+\dfrac{10}{5}
\\\\
y=-\dfrac{2}{5}x+2
.\end{array}
Hence, the slope of this line is $
-\dfrac{2}{5}
.$
Since the line passing through the given point, $(
4,1
),$ is parallel to the previous line, then the slopes of these lines are equal. Using $y-y_1=m(x-x_1)$ or the point-slope form, the equation of the line is
\begin{array}{l}\require{cancel}
y-1=-\dfrac{2}{5}(x-4)
.\end{array}
In $y=mx+b$ form, the equation above is equivalent to
\begin{array}{l}\require{cancel}
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{array}
In $Ax+By=C$ form, the equation above is equivalent to
\begin{array}{l}\require{cancel}
5(y)=5\left( -\dfrac{2}{5}x+\dfrac{13}{5} \right)
\\\\
5y=-2x+13
\\\\
2x+5y=13
.\end{array}
Hence, the different forms of the equation of the line with the given conditions are
\begin{array}{l}\require{cancel}
\text{slope-intercept form: }
y=-\dfrac{2}{5}x+\dfrac{13}{5}
\\\\
\text{standard form: }
2x+5y=13
.\end{array}