Question

Use cross products to find the area of the triangle in the xy-plane defined by (1, 2), (3, 4), and (−7, 7).

Answer 1

I love these. It's often called the Shoelace Formula. It actually works for the area of any 2D polygon.

We can derive it by first imagining our triangle in the first quadrant, one vertex at the origin, one at (a,b), one at (c,d), with (0,0),(a,b),(c,d) in counterclockwise order.

Our triangle is inscribed in the $a \times d$ rectangle. There are three right triangles in that rectangle that aren't part of our triangle. When we subtract the area of the right triangles from the area of the rectangle we're left with the area S of our triangle.

$S = ad - \frac 1 2 ab - \frac 1 2 cd - \frac 1 2 (a-c)(d-b) = \frac 1 2(2 ad - ab -cd - ad +ab +cd -bc) = \frac 1 2(ad -bc)$

That's the cross product in the purest form. When we're away from the origin, a arbitrary triangle with vertices $A(x_1, y_1), B(x_2, y_2), C(x_3, y_3)$ will have the same area as one whose vertex C is translated to the origin.

We set $a=x_1 - x_3, b= y_1 - y_3, c=x_2 - x_3, d=y_2- y_3$

$S= ad-bc=(x_1 - x_3)(y_2 - y_3) -(x_2-x_3)(y_1 - y_3)$

That's a perfectly useful formula right there. But it's usually multiplied out:

$S= x_1y_2 - x_1 y_3 - x_3y_2 + x_3 y_3 - x_2 y_1 + x_2y_3 + x_3 y_1 - x_3 y_3$

$S= x_1 y_2 - x_2 y_1 + x_2y_3 - x_3y_2 + x_3 y_1 - x_1 y_3$

That's the usual form, the sum of cross products. Let's line up our numbers to make it easier.

(1, 2), (3, 4), (−7, 7)

(−7, 7),(1, 2), (3, 4),

[tex]A = \frac 1 2 ( 1(7)-2(-7) + 3(2)-4(1) + -7(4) - (7)(3)