Finding the Area of a Triangle
Here is a typical problem in analytic geometry: You are given the coordinates of two points, P1(x1,y1) and P2(x2,y2), and asked to find the area of the triangle OP1P2, where O is the origin. To keep things straightforward, I'll stipulate that P1 and P2 are in the first quadrant, and that P1 is to the left of P2. Let r1 and r2 be the lengths of OP1 and OP2, respectively, and let q be the angle between the lines OP1 and OP2 (see Figure 2).
Recall that the area of a triangle is one-half the base times the height, and, after perhaps brushing off a little mental dust, that the height h at angle q is given by h=r1sin(q). Example 2(a) gives the area of triangle OP1P2.
What remains is to somehow work the coordinates of the points P1 and P2 into this formula. This is done using a bit of trigonometry.
If a is the angle that OP1 makes with the positive x-axis, and 
is the angle that OP2 makes with the positive x-axis, then q=
-. By the definition of sine and cosine, you then have Example 2(b).
Using the identity sin(a-b)=sin(a)cos(b)- cos(a)sin(b), you find Example 2(c). And that's it. You are done! By multiplying both sides of the equation in Example 2(c) by r1r2, you arrive at Example 2(d).
— T.G.
