Question

Points ${\text{(1,1),( - 2,7),(3, - 3)}}$
The vertices of the triangles are . The it’s area is

$$A.)\;\;\;0{\text{ }}sq.{\text{ }}unit \\\ B.)\;\;\;2{\text{ }}sq.{\text{ }}unit \\\ C.)\;\;\;24{\text{ }}sq.{\text{ }}unit \\\ D.)\;\;\;12{\text{ }}sq.{\text{ }}unit \\\$$

Answer 1

As all the vertices are mentioned ${\text{(1,1),( - 2,7),(3, - 3)}}$ so using the determinant method to find the area of the triangle which is\left| {\dfrac{{\text{1}}}{{\text{2}}}\left| {\begin{array}{*{20}{c}} {{{\text{x}}_{\text{1}}}}&{{{\text{y}}_{\text{1}}}}&{\text{1}} \\\ {{{\text{x}}_{\text{2}}}}&{{{\text{y}}_{\text{2}}}}&{\text{1}} \\\ {{{\text{x}}_{\text{3}}}}&{{{\text{y}}_{\text{3}}}}&{\text{1}} \end{array}} \right|} \right|, we simply put the values and the answer will be obtained.

Complete step-by-step answer:
So we can use the above provided info as ,
Just we need to substitute the values of the vertices ${\text{(1,1),( - 2,7),(3, - 3)}}$ in the determinant given as \left| {\dfrac{{\text{1}}}{{\text{2}}}\left| {\begin{array}{*{20}{c}} {{{\text{x}}_{\text{1}}}}&{{{\text{y}}_{\text{1}}}}&{\text{1}} \\\ {{{\text{x}}_{\text{2}}}}&{{{\text{y}}_{\text{2}}}}&{\text{1}} \\\ {{{\text{x}}_{\text{3}}}}&{{{\text{y}}_{\text{3}}}}&{\text{1}} \end{array}} \right|} \right|
And it can be simplified to and solved further as

\Rightarrow \left| {\dfrac{1}{2}\left| {\begin{array}{*{20}{c}} 1&{\text{1}}&{\text{1}} \\\ {{\text{ - 2}}}&{\text{7}}&{\text{1}} \\\ 3&{ - 3}&{\text{1}} \end{array}} \right|} \right| \\\ \Rightarrow \left| {\dfrac{1}{2}[1(7 - ( - 3)) - 1( - 2 - 3) + 1(6 - 21)]} \right| \\\ \Rightarrow \left| {\dfrac{1}{2}[10 + 5 - 15]} \right| \\\ \Rightarrow 0sq.unit \\\

Hence , 0 sq. unit is our answer and so option (a) is the required answer.

Note: Triangle : A triangle is a polygon with three edges and three vertices. It is one of the basic shapes in geometry. A triangle with vertices A, B, and C is denoted. And also if the area of the triangle is zero then it states that all the points are collinear.
And also if in 3-D we need to find the area of triangle than it’s formula is given as \left| {\dfrac{{\text{1}}}{{\text{2}}}\left| {\begin{array}{*{20}{c}} {{{\text{x}}_{\text{1}}}}&{{{\text{y}}_{\text{1}}}}&{{{\text{z}}_{\text{1}}}} \\\ {{{\text{x}}_{\text{2}}}}&{{{\text{y}}_{\text{2}}}}&{{{\text{z}}_{\text{2}}}} \\\ {{{\text{x}}_{\text{3}}}}&{{{\text{y}}_{\text{3}}}}&{{{\text{z}}_{\text{3}}}} \end{array}} \right|} \right| for the vertices $({x_{\text{1}}}{\text{,}}{{\text{y}}_{\text{1}}},{z_{\text{1}}}),({x_2}{\text{,}}{{\text{y}}_2},{z_2}),({x_3}{\text{,}}{{\text{y}}_3},{z_3})$