Consider the triangle with vertices $(1,2)$, $(-5,-1)$ and $(3,-2)$. Let $\triangle$ denote the region enclosed by the above triangle. Consider the function $f:\triangle\to \mathbb{R}$ denoted by $f(x,y)=|10x-3y|$. Then the range of $f$ is the interval $$(a)~[0,36];\quad(b)~[0,47];\quad(c)~[4,47];\quad(d)~[36,47];$$
I just draw the triangle and find $f(x,y)$ for arbitrarily values of $(x,y)$ in $\triangle$ and try to discard one by one option. For example, the point $(0,0)$ is in $\triangle$ and $f(0,0)=0.$ So the option $(c)$ and $(d)$ are not correct. But I'm stuck for option $(a)$ and $(b)$.
Is their any direct method and how can I ensure that the range is an interval without seeing the option.