$\begin{array}{1 1}(A)\;\text{isosceles and right angled} \\(B)\;\text{isosceles but not right angled}\\(C)\;\text{right angled but not isosceles } \\(D)\;\text{ neither right angled non isosceles} \end{array}$

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$AB = \sqrt {(4+1)^2+(0+1)^2}$

$\qquad= \sqrt {(5)^2+(1)^2}$

$\qquad= \sqrt {26}$

$BC = \sqrt {(3+1)^2+(5+1)^2}$

$\qquad= \sqrt {(4)^2+(6)^2}$

$\qquad= \sqrt {52}$

$CA = \sqrt {(4-3)^2+(0-5)^2}$

$\qquad= \sqrt {(1)^2+(5)^2}$

$\qquad= \sqrt {26}$

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.

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