If $x-2y+4=0$   and $2x+y-5=0$ are the sides of an isosceles triangle having area 10sq.units, then possible equation of the third side is

Given lines are mutually perpendicular and intersect at  $A(\frac{6}{5},\frac{13}{5})$. 

Equation of angle bisectors  $B_{1}and{B}_2$ of the given lines are  $x-2y+4=\pm (2x+y-5)$
or  $x+3y=9and3x-y=1$
Side BC will be parallel to these bisectors.
Let  $AD=a$.
$\therefore AB=a\sqrt{2}$ 
Area of triangle  $ABC=\frac{1}{2}\times {\left(a\sqrt{2}\right)}^2=a^2$
$\therefore a^2=10$ (given)
or  $a=\sqrt{10}$
Let equation of BC be $x+3y=\lambda$ . Then
$\sqrt{10}=\frac{|\frac{6}{5}+\frac{39}{5}-\lambda |}{\sqrt{10}}$ 
 $\lambda =-1,19$
So, equation of BC is  $x+3y=-1orx+3y=19$
If equation of BC is $3x-y=\lambda$  , then
 $\sqrt{10}=\frac{|\frac{18}{5}-\frac{13}{5}-\lambda |}{\sqrt{10}}$
 $\lambda =-9,11$
Therefore, equation of BC is  $3x-y=-9or3x-y=11$