Very simple triangle area problem I have a problem with a triangle question. The question is: Imagine a triangle with the points $A(-5, 0), B(-3,-7), C(2,-2)$, where the sides $AB$ and $AC$ are equal. What is the area of this triangle?
My solution is simply to draw a bounding box around the triangle with the base $ 2-(-5) = 7$ and the height of $7$ ($-7$ from $B$). This gives me the area of $\frac{49}{2}$. The answer should be $\frac{45}{2}$ according to my text book, what am I doing wrong?
Thank you.
 A: 
$$7^2 - \frac{7\times 2}{ 2} -\frac{ 5 \times 5}{ 2} -\frac{2 \times 7 }{ 2} = \frac{45}{2}$$
This will be less than half the rectangle unless two of the vertices are at corners.  
Method 5 here shows the calculation should be $\dfrac{7\times 7}{ 2} -\dfrac{ 2 \times 2}{ 2}$ or more generally $\dfrac{fg}{ 2} -\dfrac{vw}{ 2}$.

A: Your approach is obviously wrong. Imagine a "collapsed triangle" $A(0,0)$, $B(10,10)$, $C(10,10)$. The area is obviously zero, but your method gives $50$.
The correct approach is to remember that length of a vector product of 2 vectors is the area of the rhomboid they form. The trangle has is half as large.
Therefore $x=B-A=(2,-7,0)$ and $y=C-A=(7,-2,0)$ (we have to consider the vectors in 3D). Then $x\times y=(0,0,45)$, it's length is obviously $45$ and therefore the area is $45/2$.
A: If you have three points $(x_a,y_a)$, $(x_b,y_b)$, and $(x_c,y_c)$ you can simply compute the (signed) triangles area by 
$$\frac{1}{2}\det\; \begin{pmatrix} x_a & x_b & x_v \\ y_a & y_b & y_v \\ 1 & 1 & 1 \end{pmatrix}.$$
In your case it is 
$$\frac{1}{2}\det\; \begin{pmatrix} -5 & -3 & 2 \\ 0 & -7 & -2 \\ 1 & 1 & 1 \end{pmatrix}=(35+6-10+14)/2=45/2.$$
The formula generalizes nicely to higher dimensions.
