Answer
The set $S=\{(4,-3),(8,-6)\}$ is not a basis for $R^2$ .
Work Step by Step
The set $S=\{(4,-3),(8,-6)\}$ is not a basis for $R^2$ because the vectors in $S$ are not linearly independent. Because one can see that $(8,-6)=2(4,-3)$, and hence we have the non trivial combination
$$2(4,-3)-(8,-6)=(0,0)$$
Which means that the vectors $(4,-3),(8,-6)$ are not Linearly independent. Hence, the set $S=\{(4,-3),(8,-6)\}$ is not a basis for $R^2$ .
