Question

What is the area of triangle ABC with vertices $A\left( {2,3} \right)$, $B\left( {1, - 3} \right)$and $C\left( { - 3,1} \right)$?

Answer 1

Hint: Assume the coordinates of vertices \[A\left( {2,3} \right),B\left( {1, - 3} \right)\,and\,C\left( { - 3,1} \right)\] as \[\left( {{x_1},{y_1}} \right),\left( {{x_2},{y_2}} \right)\,and\,\left( {{x_3},{y_3}} \right)\] respectively. Apply the formula for area of triangle in coordinate form given as: \[Area = \dfrac{1}{2}[{x_1}({y_2} - {y_3}) + {x_2}({y_3} - {y_1}) + {x_3}({y_1} - {y_2})]\] . If the area turns out to be positive then that will be the required answer but if it turns out to be negative then take modulus so that it becomes positive.

Complete step by step solution:
Here, we have been provided with a triangle ABC whose vertices are given as \[A\left( {2,3} \right),B\left( {1, - 3} \right)\,and\,C\left( { - 3,1} \right)\].
Now, assuming the coordinates of A, B and C as \[\left( {{x_1},{y_1}} \right),\left( {{x_2},{y_2}} \right)\,and\,\left( {{x_3},{y_3}} \right)\], we get
\[A(2,3) = ({x_1},{y_1})\,\,,\,\,B(1, - 3) = ({x_2},{y_2})\,\,,\,\,C( - 3,1) = ({x_3},{y_3})\]
Now, we know that area of triangle with coordinates \[\left( {{x_1},{y_1}} \right),\left( {{x_2},{y_2}} \right)\,and\,\left( {{x_3},{y_3}} \right)\] is given by the expression:
\[Area = \dfrac{1}{2}[{x_1}({y_2} - {y_3}) + {x_2}({y_3} - {y_1}) + {x_3}({y_1} - {y_2})]\]
Now, substituting the values of \[{x_1},{x_2},{x_3},{y_1},{y_2}\,and\,{y_3}\] in the above expression for area, we get,
\[Ar.(\Delta ABC) = \dfrac{1}{2}[2( - 3 - 1) + 1(1 - 3) + \left( { - 3} \right)(3 - \left( { - 3} \right))]\]
\[ \Rightarrow Ar.(\Delta ABC) = \dfrac{1}{2}[2( - 4) + 1( - 2) + \left( { - 3} \right)(6)]\]
\[ \Rightarrow Ar.(\Delta ABC) = \dfrac{1}{2}[ - 8 + \left( { - 2} \right) + \left( { - 18} \right)]\]
\[ \Rightarrow Ar.(\Delta ABC) = \dfrac{1}{2}[ - 28]\]
\[ \Rightarrow Ar.(\Delta ABC) = - 14\]
As we can see that the area of the \[\Delta ABC\] turns out to be a negative value and we know that area cannot be negative. So, we must take the modulus on both sides to get the value of the area positive. Therefore, taking modulus, we get,
\[\therefore Ar.\left( \Delta ABC \right) = \left| { - 14} \right| = 14{\text{square units}}\]
Hence, the area of the \[\Delta ABC\] is $ 14$ square units.

Note:
There is one more method to solve this question. We can also use the heron’s formula to answer this question. For that first we have to find the length of each side using the two vertices of the triangle by applying the distance formula $d = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} $. Then, we have to use the heron’s formula of area $Area = \sqrt {s\left( {s - a} \right)\left( {s - b} \right)\left( {s - c} \right)} $ where a, b, c, are the sides the triangle and $s = \dfrac{{a + b + c}}{2}$ to find the answer of this question.