Statement

Let v1,v2R2v_1,v_2\in\mathbb R^2 be linearly independent and H=[v1v2]H=[v_1\mid v_2]. Then

Tpos{v1,v2}H1T[0,)2.T\in\operatorname{pos}\{v_1,v_2\} \quad\Longleftrightarrow\quad H^{-1}T\in[0,\infty)^2.

Membership in the interior is equivalent to both coefficients being strictly positive. Thus the test is an ordinary matrix inverse followed by componentwise inequalities.

Oriented determinant form

If det(v1,v2)>0\det(v_1,v_2)>0, the coefficients are

y1=det(T,v2)det(v1,v2),y2=det(v1,T)det(v1,v2).y_1=\frac{\det(T,v_2)}{\det(v_1,v_2)},\qquad y_2=\frac{\det(v_1,T)}{\det(v_1,v_2)}.

Their nonnegativity tests which side of each boundary ray contains TT. If the determinant is negative, its sign must be retained in these formulas.