If the vertex of a parabola is (4, 3) and its directrix is $3x+2y-7=0,$then the equation of latus rectum of the parabola is 

Let $A=\left(4, 3\right),$ S be the focus.

Let $Z=\left(h, k\right)$ be the foot of the perpendicular from A to the directrix $3x+2y-7=0$

$\therefore$$\frac{h-4}{3}=\frac{k-3}{2}=-\frac{\left(12+6-7\right)}{9+4}\Rightarrow \frac{h-4}{3}=\frac{k-3}{2}=\frac{-11}{13}\Rightarrow h=\frac{19}{13}, k=\frac{17}{13}$$\Rightarrow Z=\left(\frac{19}{13}, \frac{17}{13}\right)$

Clearly A is the midpoint of SZ.     $\therefore$$A=\frac{Z+S}{2}$

$\Rightarrow S=2A-Z\Rightarrow S=\left(8-\frac{19}{13}, 6-\frac{17}{13}\right)\Rightarrow S=\left(\frac{85}{13},\frac{61}{13}\right)$

Let the equation of the latus rectum be $3x+2y=k$

Since it passes through S, we have $3\left(\frac{89}{13}\right)+2\left(\frac{61}{13}\right)=k.$        $\therefore k=29$

Equation of Latus rectum of the parabola is $3x+2y-29=0$.