Answer
The function of the graph perpendicular to the line $2x-5y-10=0$ and passing through the point $\left( {{x}_{1}},\ {{y}_{1}} \right)=\left( -4,\ -3 \right)$ is $y=-\frac{5}{2}x-13$.
Work Step by Step
The equation of the line is $y-{{y}_{1}}=m\left( x-{{x}_{1}} \right)$ ,
Here, the point $\left( {{x}_{1}},\ {{y}_{1}} \right)=\left( -4,\ -3 \right)$.
The value of the slope $m=\frac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}$.
It is given that the equation is perpendicular to the line.
Thus,
$\begin{align}
& 2x-5y-10=0 \\
& y=\frac{1}{5}\left( 2x-10 \right) \\
& =\frac{2}{5}x-2
\end{align}$
Now, the slope of the equation is
$\begin{align}
& m=-\frac{1}{\left( \frac{2}{5} \right)} \\
& =-\frac{5}{2}
\end{align}$.
So, the graph is passing through the point $\left( {{x}_{1}},\ {{y}_{1}} \right)=\left( -4,\ -3 \right)$
Thus:
$\begin{align}
& y-{{y}_{1}}=m\left( x-{{x}_{1}} \right) \\
& y+3=-\frac{5}{2}\left( x+4 \right) \\
& y=-\frac{5}{2}x-\frac{20}{2}-3 \\
& =-\frac{5}{2}x-13
\end{align}$
