Find the area of a triangle formed by the points A(6, 2), B(4, 7) and C(7, −4)
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 3.5 unit2
The area of a triangle with vertices \((x_1,y_1)\), \((x_2,y_2)\), \((x_3,y_3)\) is \(\frac{1}{2}\left|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right|\); the absolute value is taken because an area cannot be negative. Here the vertices are \((6,2)\), \((4,7)\) and \((7,-4)\):
First term, from A: \(6(7-(-4))=6\times11=66\)
Second term, from B: \(4(-4-2)=4\times(-6)=-24\)
Third term, from C: \(7(2-7)=7\times(-5)=-35\)
Half the size of their sum: \(\displaystyle\text{Area}=\frac{1}{2}\left|66-24-35\right|=\frac{7}{2}=3.5\) square units