#### Answer

$\text{a) Slope-Intercept Form: }
y=-\dfrac{5}{2}x
\\\text{b) Standard Form: }
5x+2y=0$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
Find the slope of the line defined by the given equation, $
2x-5y=6
.$ Then use the Point-Slope Form of linear equations with the given point, $
(0,0)
$ to find the equation of the needed line. Finally, express the answer in both the Slope-Intercept and Standard forms.
$\bf{\text{Solution Details:}}$
In the form $y=mx+b,$ the given equation is equivalent to
\begin{array}{l}\require{cancel}
2x-5y=6
\\\\
-5y=-2x+6
\\\\
\dfrac{-5y}{-5}=\dfrac{-2x}{-5}+\dfrac{6}{-5}
\\\\
y=\dfrac{2}{5}x-\dfrac{6}{5}
.\end{array}
Using $y=mx+b$ or the Slope-Intercept Form, where $m$ is the slope, then the slope of the line with the equation above is
\begin{array}{l}\require{cancel}
m=\dfrac{2}{5}
.\end{array}
Since perpendicular lines have negative reciprocal slopes, then the needed line has the following characteristics:
\begin{array}{l}\require{cancel}
m=-\dfrac{5}{2}
\\\text{Through: }
(0,0)
.\end{array}
Using $y-y_1=m(x-x_1)$ or the Point-Slope Form of linear equations, the equation of the line with the given conditions,
\begin{array}{l}\require{cancel}
y_1=
0
,\\x_1=
0
,\\m=
-\dfrac{5}{2}
,\end{array}
is
\begin{array}{l}\require{cancel}
y-y_1=m(x-x_1)
\\\\
y-0=-\dfrac{5}{2}(x-0)
\\\\
y=-\dfrac{5}{2}x
.\end{array}
In the form $y=mx+b$ or the Slope-Intercept Form, the equation above is equivalent to
\begin{array}{l}\require{cancel}
y=-\dfrac{5}{2}x
.\end{array}
In the form $ax+by=c$ or the Standard Form, the equation above is equivalent to
\begin{array}{l}\require{cancel}
2(y)=2\left( -\dfrac{5}{2}x \right)
\\\\
2y=-5x
\\\\
5x+2y=0
.\end{array}
Hence, $
\text{a) Slope-Intercept Form: }
y=-\dfrac{5}{2}x
\\\text{b) Standard Form: }
5x+2y=0
.$