Area of a Triangle - what is the formula?

Question

Triangle $ABC$: $A(1; 4), ~B(7; 8), ~C(9; 2)$.

I need to find the area of a triangle. I found this solution:

But, what is the formula?

Answer 1

If the vertices are $(0,0), (x_1, y_1), (x_2, y_2)$, the area is given by $\frac{1}{2}|x_1y_2 - x_2y_1|$. Now, transferring the origin to $A$, the coordinates of the vertices become $(x_B-x_A, y_B-y_A), (x_C-x_A, y_C-y_A), (0,0)$ and hence the area is $$\frac{1}{2}|(x_B-x_A)(y_C-y_A) - (x_C-x_A)(y_B-y_A)|$$

Answer 2

This formula is a simplified form of this: ‎\begin{eqnarray*}‎ ‎S=\frac12‎ ‎\Big[‎ ‎ \left|\begin{array}{rr}x_1&x_2\\y_1&y_2\end{array}\right|‎ + ‎ \left|\begin{array}{rr}x_2&x_3\\y_2&y_3\end{array}\right|‎ ‎ %+\left|\begin{array}{rr}x_3&x_4\\y_3&y_4\end{array}\right|‎ + ‎ \cdots‎ + ‎ \left|\begin{array}{rr}x_{n-1}&x_n\\y_{n-1}&y_n\end{array}\right|‎ + ‎ \left|\begin{array}{rr}x_n&x_1\\y_n&y_1\end{array}\right|‎ ‎\Big]‎ ‎\end{eqnarray*}‎ Where $n$ points ‎$$ P_1\Big|\begin{array}{r}x_1\\y_1\end{array},~~P_2\Big|\begin{array}{r}x_2\\y_2\end{array},~~P_3\Big|\begin{array}{r}x_3\\y_3\end{array},~~\cdots~~,~~P_n\Big|\begin{array}{r}x_n\\y_n\end{array} $$‎ are vertices of a plane polygon. For your points $A(1,4), ~B(7,8), ~C(9,2)$: ‎\begin{eqnarray*}‎ ‎S=\frac12‎ ‎\Big[‎ ‎ \left|\begin{array}{rr}1&7\\4&8\end{array}\right|‎ + ‎ \left|\begin{array}{rr}7&9\\8&2\end{array}\right|‎ ‎ +\left|\begin{array}{rr}9&1\\2&4\end{array}\right|‎ ‎\Big]‎=\frac12|8-28+14-72+36-2|=22 ‎\end{eqnarray*}‎

Answer 3

This is use of determinant to find area.

$$\frac 12 \left|\begin{array}{ccc} 1 & 1 & 1 \\ x_A & x_B & x_C \\ y_A & y_B & y_C \end{array} \right| $$

Operate Column 2 - Column 1 and Column 3 - Column 1. You get,

$$\frac 12 \left|\begin{array}{ccc} 1 & 0 & 0 \\ x_A & x_B - x_A & x_C - x_A \\ y_A & y_B - y_A & y_C - y_A \end{array} \right| $$

Solve it you got above formula.