Question

How do you find the area of a triangle whose vertices of triangle are A$\left(2,-4\right)$, B$\left(1,3\right)$, C$\left(-2,-1\right)$?

Answer 1

Hint: In order to determine the area of $\mathrm{\Delta }ABC$,you can clearly see that the coordinates are in the 2-Dimensional plane ,so directly use the formula for Area of$\mathrm{\Delta }ABC$equal to $\left({\displaystyle \frac{1}{2}}\right)\left[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right]$.The unit for area of triangle will be square units.

Formula Used:
Area of $\mathrm{\Delta }ABC$$=\left({\displaystyle \frac{1}{2}}\right)\left[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right]$

Complete step-by-step solution:
Given a triangle let it be $\mathrm{\Delta }ABC$having vertices as A$\left(2,-4\right)$, B$\left(1,3\right)$, C$\left(-2,-1\right)$
Since, we are given the vertices in 2-dimensional plane,
So in order to calculate the area of triangle having vertices in 2-dimensional plane having vertices as are A$\left(x_1,y_1\right)$, B$\left(x_2,y_2\right)$, C$\left(x_3,y_3\right)$as
Area of $\mathrm{\Delta }ABC$$=\left({\displaystyle \frac{1}{2}}\right)\left[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right]$
Putting the value of coordinates into the formula ,
Area of $\mathrm{\Delta }ABC$=
$$\begin{gathered} \Rightarrow \left({\displaystyle \frac{1}{2}}\right)\left[2(3-\left(-1\right))+1(-1-\left(-4\right))+\left(-2\right)(-4-3)\right] \\ \Rightarrow \left({\displaystyle \frac{1}{2}}\right)\left[2(3+1)+1(-1+4)+\left(-2\right)(-7)\right] \\ \Rightarrow \left({\displaystyle \frac{1}{2}}\right)\left[2(4)+1(3)+14\right] \\ \Rightarrow \left({\displaystyle \frac{1}{2}}\right)\left[8+3+14\right] \\ \Rightarrow \left({\displaystyle \frac{1}{2}}\right)\left[25\right] \\ \Rightarrow {\displaystyle \frac{25}{2}} \\ \Rightarrow 12.5sq.units \end{gathered}$$
Therefore, Area of $\mathrm{\Delta }ABC$is equal to $$12.5sq.units$$

Note:
1. A triangle is a closed geometric shape having three no of edges and three vertices. A triangle having vertices named as A,B and C is denoted by $\mathrm{\Delta }ABC$.
2.Area of Triangle in a 2-dimensional plane can be determined by following ways depending upon what type of information is given .
If length of base and height is given then
Area of $\mathrm{\Delta }ABC$=${\displaystyle \frac{1}{2}}\left(base\times height\right)$
If length of all the sides are given then,
Using heron’s formula
$s={\displaystyle \frac{a+b+c}{2}}$
Area of $\mathrm{\Delta }ABC$=$\sqrt{s(s-a)(s-b)(s-c)}$
If coordinates of the all 3 vertices in cartesian plane are given then,
Area of $\mathrm{\Delta }ABC$$=\left({\displaystyle \frac{1}{2}}\right)\left[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right]$