Consider a triangle with vertices at (0,0) and (x1, 71) and (x2, 92). a. Use the interpretation of a determinant as area of a parallelogram to explain why the following formula calculates the area of the triangle. (Hint: A diagonal of a parallelogram cuts the parallelogram into two triangles of equal size A =\det (i 2) b. Set (x1, y.) = (1,2) and (x2,Y2) = (3,-1). Use the formula to calculate the area of the triangle.

I can't really read the formula so I'll give a lecture.With two vectors we can make a parallelogram.  If the vectors are u and v, and one vertex is the origin O, the others are O+u, O+v, and O+v+u.   If we draw any diagonal of the parallelogram, that divides it into two congruent triangles.  So each triangle is half the area of parallelogram.The signed area of the parallelogram is given by the two D cross product of the vectors, aka the determinant.   So the area of a triangle with vertices (0,0), (a,b) and (c,d) is[tex]A = \frac 1 2 |ad - bc| = \left| \frac 1 2 \begin{vmatrix} a & b \\ c & d\end{vmatrix} \right|[/tex]Applying that to vertices[tex](0,0), (x_1, 71), (x_2, 92)[/tex]we get area[tex]A = \frac 1 2 |\ 92x_1 - 71x_2|[/tex]That formula is accurate for any values of x₁ and x₂b. (1,2),(3,-1)[tex]A=\frac 1 2|1(-1)-2(3)| = \frac 1 2 |-7| = \frac 7 2[/tex]Answer: 7/2