The base vertices of a right angled isosceles triangle are $(2, 4)$ and $(4, 2)$ then its third vertex is

ABC be a right angle isosceles triangle.

Extremities of hypotenuse of triangle are A=(2, 4), B=(4, 2)

Length of hypotenuse 

$$\begin{gathered} \mathrm{AB}=\sqrt{{(4-2)}^2+{(2-4)}^2} \\ =\sqrt{4+4} \\ =2\sqrt{2} \end{gathered}$$

 Since it is an isosceles triangle, let us assume its height and base = a

from ${∆}^{\mathrm{le}} \mathrm{ACB}, {\mathrm{AC}}^2+{\mathrm{BC}}^2={\mathrm{AB}}^2$

 $$\begin{gathered} \Rightarrow {\mathrm{a}}^2+{\mathrm{a}}^2=8 \\ \Rightarrow 2{\mathrm{a}}^2=8 \\ \Rightarrow {\mathrm{a}}^2=4 \Rightarrow \mathrm{a}=2 \end{gathered}$$

By verifying options, first option only satisfies the condition AC = BC = 2