Chapter 10 - Section 10.1 - The Rectangular Coordinate System - Exercises - Page 457: 35

9

First of all, we see that point C is $(5,-3)$, for it is B reflected across the x axis. Now, we can break this into triangles, one below the point (2,1) and one above the point (2,1). By doing this, we can use the distance between (2,1) and (5,3) as the base of each triangle, allowing us to use whole numbers. $A_1 = (1/2)(3)(4) = 6$ $A_2 = (1/2)(3)(2) = 3$ Thus: $A= 3+6 =9$