Answer
$y=m(x+2)+\dfrac{1}{3}$
Work Step by Step
First, we determine the intersection point of the given lines:
$\begin{cases}
5x-3y+11=0\\
2x-9y+7=0
\end{cases}$
$\begin{cases}
5x+11=3y\\
2x+7=9y
\end{cases}$
$\begin{cases}
y=\dfrac{5x+11}{3}\\
y=\dfrac{2x+7}{9}
\end{cases}$
$\dfrac{5x+11}{3}=\dfrac{2x+7}{9}$
$3(5x+11)=2x+7$
$15x+33=2x+7$
$15x-2x=7-33$
$13x=-26$
$x=\frac{-26}{13}$
$x=-2$
$y=\dfrac{5(-2)+11}{3}$
$y=\dfrac{1}{3}$
The point is: $\left(-2,\dfrac{1}{3}\right)$
The equation of the lines passing through this point is:
$y-\dfrac{1}{3}=m(x+2)$
$y=m(x+2)+\dfrac{1}{3}$
