Answer
The set $S=\{(-1,2),(1,-2),(2,4)\}$ is not a basis for $R^2$.
Work Step by Step
The set $S=\{(-1,2),(1,-2),(2,4)\}$ is not a basis for $R^2$ because the vectors in $S$ are not linearly independent. Indeed, considering the linear combination $$a (-1,2)+b(1,-2)+c(2,4)=(0,0),$$
then we have the system
$$\begin{align*}
-a+b+2c&=0\\
2a-2b+4c&=0
\end{align*}$$
It is easy to see that the above system has the solution
$a=t, b=t , c=0$ . Hence, the vectors of $S$ are not linearly independent. Hence, the set $S=\{(-1,2),(1,-2),(2,4)\}$ is not a basis for $R^2$
