The joint equation of the bisectors of the angles between the line $x=5$ and $\mathrm{y}=3$ is

Lines $x=5$ and $y=3$ are rewritten as $L_1: x-5=0$ and $L_2: y-3=0$. The angle bisectors of two lines $A_1x + B_1y + C_1 = 0$ and $A_2x + B_2y + C_2 = 0$ are given by

$\frac{A_1x+B_1y+C_1}{\sqrt{A_1^2+B_1^2}} = \pm \frac{A_2x+B_2y+C_2}{\sqrt{A_2^2+B_2^2}}$

With $A_1=1$, $B_1=0$, $C_1=-5$ and $A_2=0$, $B_2=1$, $C_2=-3$, the denominators are both 1, so the equation simplifies to $x-5 = \pm (y-3)$.

This yields two bisectors: $x-y-2=0$ and $x+y-8=0$. Their product forms the joint equation

$(x-y-2)(x+y-8)=0$

Expanding gives $x^2 - y^2 - 10x + 6y + 16 = 0$, which matches option A.

$\boxed{A}$