Question

If the vertices of a triangle are \[\left( {1,1} \right),\left( { - 2,7} \right)\] and \[\left( {3, - 3} \right)\], then its area is
A) 0 sq. units
B) 2 sq. units
C) 24 sq. units
D) 12 sq. units

Answer 1

Hint:
Here, we are required to find the area of the triangle whose all the three vertices are given. We will use the formula of area of a triangle and substitute the given vertices in that formula to find the required area. By this, we will get the required answer.

Formula Used:
Area of a triangle \[ = \left| {\dfrac{1}{2}\left[ {{x_1}\left( {{y_2} - {y_3}} \right) + {x_2}\left( {{y_3} - {y_1}} \right) + {x_3}\left( {{y_1} - {y_2}} \right)} \right]} \right|\]

Complete step by step solution:
Let the vertices of the triangle be \[A = \left( {1,1} \right)\], \[B = \left( { - 2,7} \right)\]and \[C = \left( {3, - 3} \right)\].
Hence, we have to find the area of the triangle \[ABC\] whose three vertices are given.
Now, substituting \[\left( {{x_1},{y_1}} \right) = \left( {1,1} \right)\], \[\left( {{x_2},{y_2}} \right) = \left( { - 2,7} \right)\] and \[\left( {{x_3},{y_3}} \right) = \left( {3, - 3} \right)\] in the formula Area of a triangle \[ = \left| {\dfrac{1}{2}\left[ {{x_1}\left( {{y_2} - {y_3}} \right) + {x_2}\left( {{y_3} - {y_1}} \right) + {x_3}\left( {{y_1} - {y_2}} \right)} \right]} \right|\], we get
Area of triangle \[ABC = \left| {\dfrac{1}{2}\left[ {1\left( {7 - \left( { - 3} \right)} \right) - 2\left( { - 3 - 1} \right) + 3\left( {1 - 7} \right)} \right]} \right|\]
Adding and subtracting the terms, we get
\[ \Rightarrow \] Area of triangle \[ABC = \left| {\dfrac{1}{2}\left[ {1\left( {10} \right) - 2\left( { - 4} \right) + 3\left( { - 6} \right)} \right]} \right|\]
Multiplying the terms, we get
\[ \Rightarrow \] Area of triangle \[ABC = \left| {\dfrac{1}{2}\left[ {10 + 8 - 18} \right]} \right|\]
Adding and subtracting the terms, we get
\[ \Rightarrow \] Area of triangle \[ABC = \left| {\dfrac{1}{2}\left[ 0 \right]} \right| = \left| 0 \right| = 0\]
Here, we have used the modulus sign because the area of the triangle cannot be negative.
Hence, Area of triangle ABC \[ = 0\] square units
Therefore, if the vertices of a triangle are \[\left( {1,1} \right),\left( { - 2,7} \right)\] and \[\left( {3, - 3} \right)\], then its area is 0 square units.

Hence, option A is the correct answer.

Note:
We can also find the area of the triangle using the help of determinants.
Area of triangle\[ = \left| {\dfrac{1}{2}\left| \begin{array}{l}{x_1}{\rm{ }}{y_1}{\rm{ }}1\\{x_2}{\rm{ }}{y_2}{\rm{ }}1\\{x_3}{\rm{ }}{y_3}{\rm{ }}1\end{array} \right|} \right|\]
Now, Substituting \[\left( {{x_1},{y_1}} \right) = \left( {1,1} \right)\], \[\left( {{x_2},{y_2}} \right) = \left( { - 2,7} \right)\] and \[\left( {{x_3},{y_3}} \right) = \left( {3, - 3} \right)\] in the above formula, we get,
Area of triangle\[ABC = \left| {\dfrac{1}{2}\left| \begin{array}{l}{\rm{ }}1{\rm{ }}1{\rm{ }}1\\ - 2{\rm{ }}7{\rm{ }}1\\{\rm{ }}3{\rm{ }} - 3{\rm{ }}1\end{array} \right|} \right|\]
Now, solving the determinant,
\[ \Rightarrow \] Area of triangle \[ABC = \left| {\dfrac{1}{2}\left[ {1\left( {7 - \left( { - 3} \right)} \right) - 1\left( { - 2 - 3} \right) + 1\left( {6 - 21} \right)} \right]} \right|\]
\[ \Rightarrow \] Area of triangle\[ABC = \left| {\dfrac{1}{2}\left[ {1\left( {10} \right) - 1\left( { - 5} \right) + 1\left( { - 15} \right)} \right]} \right|\]
Solving the above equation further, we get
\[ \Rightarrow \] Area of triangle\[ABC = \left| {\dfrac{1}{2}\left[ {10 + 5 - 15} \right]} \right| = \left| {\dfrac{0}{2}} \right| = 0\]
Therefore, area of triangle ABC \[ = 0\] square units
Hence, option A is the correct answer.
Also, we have used the ‘modulus sign’ while finding the area of the triangle because it means that we have to take the absolute value of the terms present inside it, i.e. we will only take the non-negative values of the terms present inside the modulus when we will remove it. Hence, we have used Modulus, keeping in mind that area of a triangle can never be negative.