Find area of the triangle ABC, given the coordinates of vertices in plane 
$A, B$ and $C$ are the points $(7,3), (-4,1)$ and $(-3,-2)$ respectively. Find the area of the triangle $ABC$.

I've worked out the lengths of each side of the triangle which are $AB=5\sqrt5$, $BC=\sqrt10$ and $AC=5\sqrt5$.
I know that the formula for the area of a triangle is $\frac12hb$ but when I checked the solutions the answer to the area of this triangle is $17\frac12$.
I do not understand how this answer is achieved.
 A: Follow-through
As you have said before, the side lengths of $\triangle ABC$ is $AB=AC=5\sqrt{5}$, $BC=\sqrt{10}$, using Heron's formula, we can compute the answer.
Heron's formula states that given side lengths $a,b,c$ of $\triangle ABC$, the area is given $$\sqrt{s(s-a)(s-b)(s-c)}\tag{1}$$
Where $s$ Is the semi perimeter. ($s=\frac {a+b+c}{2}$).
So in your case, we have $$a=5\sqrt{5},b=5\sqrt{5},c=\sqrt{10}\tag{2}$$
The semi perimeter is $$\frac {10\sqrt{5}+\sqrt{10}}{2}\tag{3}$$ and plugging in the values, we have $$\sqrt{\frac {10\sqrt{5}+\sqrt{10}}{2}\left(\frac {10\sqrt{5}+\sqrt{10}}{2}-5\sqrt{5}\right)\left(\frac {10\sqrt{5}+\sqrt{10}}{2}-5\sqrt{5}\right)\left(\frac {10\sqrt{5}+\sqrt{10}}{2}-\sqrt{10}\right)}=\boxed{17.5}\tag{4}$$
A: Method-1
$$\Delta= \frac12\begin{vmatrix}
x_1 & y_1 &   1\\ 
x_2& y_2 &  1\\ 
 x_3& y_3 &  1
\end{vmatrix}$$
Method-2
This can also be used to find the area of polygon.
$$\Delta= \frac12\begin{vmatrix}
x_1 & y_1 \\ 
x_2& y_2 \\ 
 x_3& y_3 \\
x_1 &y_1
\end{vmatrix}=\frac12\left ((x_1y_2+x_2y_3+x_3y_1)-(x_2y_1+x_3y_2+x_1y_3)\right)$$
A: You can use $BC$ as the base and measure the height with respect to it as the distance from a point to a line. If the line has equation $ax+by+c=0$, then the distance from the point $(x_0,y_0)$ to the line is
$$
\frac{|ax_0+by_0+c|}{\sqrt{a^2+b^2}}
$$
The line $BC$ has equation
$$
\frac{y-1}{-2-1}=\frac{x+4}{-3+4}
$$
or, in simplified form,
$$
3x+y+11=0
$$
The distance from $A$ to the line $BC$ is
$$
\frac{|3\cdot7+3+11|}{\sqrt{3^2+1^2}}=\frac{35}{\sqrt{10}}
$$
Thus the area is
$$
\frac{1}{2}\cdot\sqrt{10}\cdot\frac{35}{\sqrt{10}}=\frac{35}{2}=17+\frac{1}{2}
$$
Actually, there is a much simpler formula when one of the points is the origin. Suppose you want to determine the area of the triangle $OBC$, where $B(x_1,y_1)$ and $C(x_2,y_2)$. The line passing through $B$ and $C$ has equation
$$
(y_1-y_2)(x-x_2)-(x_1-x_2)(y-y_2)=0
$$
or, in simplified form,
$$
(y_1-y_2)x-(x_1-x_2)y+x_1y_2-x_2y_1=0
$$
By the above formula, the distance from $O$ to the line is obtained by using $x_0=0$ and $y_0=0$, so it is
$$
\frac{|x_1y_2-x_2y_1|}{\sqrt{(y_1-y_2)^2+(x_1-x_2)^2}}
$$
and at the denominator you recognize the length of $BC$, so the area is
$$
\frac{1}{2}|x_1y_2-x_2y_1|=
\frac{1}{2}
\left|
\det\begin{bmatrix} x_1 & x_2 \\ y_1 & y_2 \end{bmatrix}
\right|
$$
Well, your triangle can be seen as having the origin as one of its vertices, by the translation mapping $A$ to the origin! The new coordinates of $B$ and $C$ in the translated frame of reference are
$$
(-4-7,1-3)=(-11,-2)
\qquad\text{and}\qquad
(-3-7,-2-3)=(-10,-5)
$$
so the area is
$$
\frac{1}{2}
\left|
\det\begin{bmatrix} -11 & -10 \\ -2 & -5 \end{bmatrix}
\right|=\frac{1}{2}\lvert-55+20\rvert=\frac{35}{2}
$$
In general, by implicitly doing the translation, the area of the triangle with vertices in $(x_0,y_0)$, $(x_1,y_1)$ and $(x_2,y_2)$ can be computed as
$$
\frac{1}{2}
\left|
\det\begin{bmatrix} x_1-x_0 & x_2-x_0 \\ y_1-y_0 & y_2-y_0 \end{bmatrix}
\right|
$$
A: When you have all three side lengths of a triangle then you can use Heron's formula to find the area:
$A=\sqrt{s(s-a)(s-b)(s-c)}$, where $s=\frac{a+b+c}{2}$
A: You considered using the formula $\text{area} = \frac{1}{2}(\text{height})(\text{base})$ here. While there are more specialized methods to this case, the approach you thought up works perfectly fine.
You just have to select a side to be the base, and compute the base and height lengths.
e.g. if you pick side $AB$ to be the base, you've already determined that it has length $5 \sqrt{5}$.
So to finish the problem, you need to find the height — that is, the distance from the line $AB$ to the point $C$. There are a number of ways to go about this, and the value turns out to be $7/\sqrt{5}$.
Thus, the area is $\frac{1}{2} \cdot (5 \sqrt{5}) \cdot \frac{7}{\sqrt{5}} = \frac{35}{2}$.
A: Since you have obtained the length of each side, using Heron's Formula is a natural way to find the area.  Let's consider the approach suomynonA suggested in the comments. Consider the figure below.

We can find the area of $\triangle ABC$ by subtracting the sum of the areas of the three right triangles $ABD$, $ACF$, and $BCE$ from the area of rectangle $ADEF$.  I will leave the details of the calculations to you.
A: I need anwer .A(6,2),B(2,4) and C(8,-4) are three in a plane in the line BC and Q is a point in the plane, such that PAQB is a rectangle. Find the (I) equation of the sides AP and BP. (ii) the coordinate of the points P (iii) the area of the rectangle
