$AB = \sqrt {(4+1)^2+(0+1)^2}$
$\qquad= \sqrt {(5)^2+(1)^2}$
$\qquad= \sqrt {26}$
$BC = \sqrt {(3+1)^2+(5+1)^2}$
$\qquad= \sqrt {(4)^2+(6)^2}$
$\qquad= \sqrt {52}$
$CA = \sqrt {(4-3)^2+(0-5)^2}$
$\qquad= \sqrt {(1)^2+(5)^2}$
$\qquad= \sqrt {26}$
In isosceles triangle side $AB=CA$
For right angled triangle $BC^2= AB^2+AC^2$
So, here $BC= \sqrt {52} $ or $BC^2= 52$
or $(\sqrt {26})^2+(\sqrt {26})^2=52$
So the given triangle is right angled and also isoceles
Hence A is the correct answer.