I have this theorem with proof that i need to understand, however the proof in the book i am into is not complete.
Thm: If $(p_1, p_2)$, $(q_1, q_2)$ and $(r_1, r_2)$ are vertices of a triangle, then the area of that triangle is the absolute value of $\frac{1}{2}[(q_1-p_1)(r_2-p_2) - (q_2-p_2)(r_1-p_1)]$.
The proof in the book says that the first shaded triangle should be proven first and it is left for the reader. The first triangle(shaded) should be proven first
Then the following is given for the proof of the second triangle,which is the goal of the proof: Since $τP,O(P)=(0,0)$, $τP,O(Q)=(a,b)$ and $τP,O(P)=(c,d)$, then the area of $∆PQR$ is also $\frac{1}{2}|ad-bc|$. Substitution then yields the expression given in the statement of the theorem.
However the only idea I have is that the translation that takes $P$ to $Q$ and $P$ to $R$ has equations: $q_1-p_1=a$ , $q_2-p_2=b$ and $r_1-p_1=c$ , $r_2-p_2=d$.
Those are simply relation between cartesian coordinates, for example with reference to PQ for the triangle on the right we have that
distance $PQ$ in $x$ direction $= q_1-p_1$
distance $PQ$ in $y$ direction $= q_2-p_2$
Since the shaded triangle is obtained by a translation of the right triangle such that $P\equiv O$, for the shaded triangle we have
distance $PQ$ in $x$ direction $a= q_1-p_1$
distance $PQ$ in $y$ direction $b = q_2-p_2$
and ths the point $Q$ for the shaded triangle has coordinates $(a,b)$.