Hello, SaRah<3!

Here's #1 . . . in baby steps . . .

1. The vertices of a right triangle are A(3,3), B(-1,4), and C(1,-5).

Find the area of the triangle.

A sketch is always a good idea . . . Code:

(-1,4)
B o |
* |
*
* | *
| *
* | *
| *
*| *
| o A
* * (3,3)
- - - + - - - * - - -
|* *
| *
| o
| C (1,-5)

Even a rough sketch makes us suspect that: $\displaystyle AB \perp AC$

We find that: .$\displaystyle m_{AB}\:=\:\frac{4-3}{-1-3}\:=\:\frac{1}{-4}\:=\:-\frac{1}{4}$

. . and that: .$\displaystyle m_{AC}\:=\:\frac{-5-3}{1-3}\:=\:\frac{-8}{-2}\:=\:+4$

Since the slopes are negative reciprocals of each other: $\displaystyle \angle A = 90^o$

The base of the triangle is: .$\displaystyle AC\;=\;\sqrt{(1-3)^2 + (-5-3)^2}\:=\:\sqrt{68}\:=\:2\sqrt{17}$

The height of the triangle is: .$\displaystyle AB\;=\;\sqrt{-1-3)^2 + (4-3)^2}\:=\;\sqrt{17}$

The area of a triangle is: .$\displaystyle A \;=\;\frac{1}{2}(base)(height)$

. . so we have: .$\displaystyle A\;=\;\frac{1}{2}\left(2\sqrt{17}\right)\left( \sqrt{17}\right) \;= \;17$ square units.

Edit: EB's too fast for me . . . and with a shorter solution!