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

Formula Used:
Area of $\Delta ABC$$ = \left( {\dfrac{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 $\Delta ABC$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 $\Delta ABC$$ = \left( {\dfrac{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 $\Delta ABC$=
\[
   \Rightarrow \left( {\dfrac{1}{2}} \right)\left[ {2(3 - \left( { - 1} \right)) + 1( - 1 - \left( { - 4} \right)) + \left( { - 2} \right)( - 4 - 3)} \right] \\
    \Rightarrow \left( {\dfrac{1}{2}} \right)\left[ {2(3 + 1) + 1( - 1 + 4) + \left( { - 2} \right)( - 7)} \right] \\
   \Rightarrow \left( {\dfrac{1}{2}} \right)\left[ {2(4) + 1(3) + 14} \right] \\
     \Rightarrow \left( {\dfrac{1}{2}} \right)\left[ {8 + 3 + 14} \right] \\
      \Rightarrow \left( {\dfrac{1}{2}} \right)\left[ {25} \right] \\
      \Rightarrow \dfrac{{25}}{2} \\
      \Rightarrow 12.5\,sq.units \\
 \]
Therefore, Area of $\Delta ABC$is equal to \[12.5\,sq.units\]

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 $\Delta ABC$.
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 $\Delta ABC$=$\dfrac{1}{2}\left( {base \times height} \right)$
If length of all the sides are given then,
Using heron’s formula
$
  s = \dfrac{{a + b + c}}{2} \\
    \\
 $
Area of $\Delta ABC$=$\sqrt {s(s - a)(s - b)(s - c)} $
If coordinates of the all 3 vertices in cartesian plane are given then,
Area of $\Delta ABC$$ = \left( {\dfrac{1}{2}} \right)\left[ {{x_1}({y_2} - {y_3}) + {x_2}({y_3} - {y_1}) + {x_3}({y_1} - {y_2})} \right]$