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Question: How do you find the area of a triangle whose vertices of triangle are A\(\left( {2, - 4} \right)\), ...

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

Explanation

Solution

In order to determine the area of ΔABC\Delta ABC,you can clearly see that the coordinates are in the 2-Dimensional plane ,so directly use the formula for Area ofΔABC\Delta ABCequal to (12)[x1(y2−y3)+x2(y3−y1)+x3(y1−y2)]\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 ΔABC\Delta ABC =(12)[x1(y2−y3)+x2(y3−y1)+x3(y1−y2)] = \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 ΔABC\Delta ABChaving vertices as A(2,−4)\left( {2, - 4} \right), B(1,3)\left( {1,3} \right), C(−2,−1)\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(x1,y1)\left( {{x_1},{y_1}} \right), B(x2,y2)\left( {{x_2},{y_2}} \right), C(x3,y3)\left( {{x_3},{y_3}} \right)as
Area of ΔABC\Delta ABC =(12)[x1(y2−y3)+x2(y3−y1)+x3(y1−y2)] = \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 ΔABC\Delta ABC=

⇒(12)[2(3−(−1))+1(−1−(−4))+(−2)(−4−3)] ⇒(12)[2(3+1)+1(−1+4)+(−2)(−7)] ⇒(12)[2(4)+1(3)+14] ⇒(12)[8+3+14] ⇒(12)[25] ⇒252 ⇒12.5 sq.units  \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 ΔABC\Delta ABCis equal to 12.5 sq.units12.5\,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 ΔABC\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 ΔABC\Delta ABC=12(base×height)\dfrac{1}{2}\left( {base \times height} \right)
If length of all the sides are given then,
Using heron’s formula
s=a+b+c2   s = \dfrac{{a + b + c}}{2} \\\ \\\
Area of ΔABC\Delta ABC=s(s−a)(s−b)(s−c)\sqrt {s(s - a)(s - b)(s - c)}
If coordinates of the all 3 vertices in cartesian plane are given then,
Area of ΔABC\Delta ABC =(12)[x1(y2−y3)+x2(y3−y1)+x3(y1−y2)] = \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]