Find area of the triangle ABC, given the coordinates of vertices in plane

Solution 1:

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.

Solution 2:

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 $$\sqrt{\frac {10\sqrt{5}+\sqrt{10}}{2}\left(\frac {10\sqrt{5}+\sqrt{10}}{2}-5\sqrt{5}\right)\left(\frac {10\sqrt{5}+\sqrt{10}}{2}-5\sqrt{5}\right)\left(\frac {10\sqrt{5}+\sqrt{10}}{2}-\sqrt{10}\right)}=\boxed{17.5}\tag{4}$$

Solution 3:

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. $$\Delta= \frac12\begin{vmatrix} x_1 & y_1 \\ x_2& y_2 \\ x_3& y_3 \\ x_1 &y_1 \end{vmatrix}=\frac12\left ((x_1y_2+x_2y_3+x_3y_1)-(x_2y_1+x_3y_2+x_1y_3)\right)$$

Solution 4:

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 $$ \frac{1}{2} \left| \det\begin{bmatrix} x_1-x_0 & x_2-x_0 \\ y_1-y_0 & y_2-y_0 \end{bmatrix} \right| $$