Answer
(a) $(a_1,b_1),(a_2,b_1),(a_2,b_3),(a_1,b_3)$
(b) and (c) see explanations.
Work Step by Step
(a) Use the figure given by the Exercise,
we can find the coordinates for the rectangle corners as $(a_1,b_1),(a_2,b_1),(a_2,b_3),(a_1,b_3)$
(b) Find the area of the red triangle by subtracting the areas of the three blue triangles from the area of the rectangle.
The area of the rectangle is $A_0=(a_2-a_1)(b_3-b_1)$
And the three blue triangles have areas of
$A_1=\frac{1}{2}(a_3-a_1)(b_3-b_1)$, (top left)
$A_2=\frac{1}{2}(a_2-a_1)(b_2-b_1)$, (bottom right)
$A_3=\frac{1}{2}(a_2-a_3)(b_3-b_2)$, (top right)
The area of the middle red triangle is then:
$A=A_0-A_1-A_2-A_3=\frac{1}{2}(a_1b_2-a_1b_3-a_2b_1+a_2b_3+a_3b_1-a_3b_2)$
Note: this step requires the expansion and combination of the $a_ib_j$ terms.
(c) We can evaluate the determinate by doing the operations $R_2-R_1\to R_2, R_3-R_1\to R_3$
$\begin{array}( \\A=\pm\frac{1}{2} \\ \\ \end{array}
\begin{vmatrix} a_1&b_1&1\\ a_2-a_1&b_2-b_1&0\\ a_3-a_1&b_3-b_1&10\end{vmatrix}$
Expand with column3 to get:
$A=\pm\frac{1}{2}[(a_2-a_1)(b_3-b_1)-(a_3-a_1)(b_2-b_1)]=\pm\frac{1}{2}(a_1b_2-a_1b_3-a_2b_1+a_2b_3+a_3b_1-a_3b_2)$
The $\pm$ signs guarantees that the result will be positive and this agrees with the result from part (b) above.
