Ans: A
Note that the slope of the straight line $9x-5y+45=0$ is $\dfrac{9}{5}$.
Note that the slope of the straight line $9x-5y+45=0$ is $\dfrac{9}{5}$.
Since $L$ and the given straight line is perpendicular to each other, then we have
$\begin{array}{rcl}
m_L & = & -1 \div \dfrac{9}{5} \\
m_L & = & \dfrac{-5}{9}
\end{array}$
The equation of $L$ is
$\begin{array}{rcl}
\dfrac{y-0}{x-(-3)} & = & \dfrac{-5}{9} \\
9y & = & -5x – 15 \\
5x +9y + 15 & = & 0
\end{array}$
