If the area of the triangle with vertices $\left( {x,0} \right),\left( {1,1} \right)$ and $\left( {0,2} \right)$ is $4$ square units, then what is the value of $x$?
A. $ - 2$
B. $ - 4$
C. $ - 6$
D. $8$

Hint: The coordinates of all the vertices of a triangle are given. One of them includes an unknown. The area of the triangle is also given. Using the formula of finding the area of a triangle you have to find the value of the unknown.

Formula Used:
Area of a triangle having coordinates of the vertices $A\left( {{x_1},{y_1}} \right),B\left( {{x_2},{y_2}} \right),C\left( {{x_3},{y_3}} \right)$ is given by $\dfrac{1}{2}\left| {{x_1}\left( {{y_2} - {y_3}} \right) + {x_2}\left( {{y_3} - {y_1}} \right) + {x_3}\left( {{y_1} - {y_2}} \right)} \right|$ square units.

Complete step by step solution:
Here the coordinates of the vertices are $\left( {x,0} \right),\left( {1,1} \right)$ and $\left( {0,2} \right)$
Let be the triangle in which $A = \left( {x,0} \right),B = \left( {1,1} \right),C = \left( {0,2} \right)$
So, ${x_1} = x,{y_1} = 0,{x_2} = 1,{y_2} = 1,{x_3} = 0,{y_3} = 2$
Substitute these values in the formula.
Area of is
$\dfrac{1}{2}\left| {x\left( {1 - 2} \right) + 1\left( {2 - 0} \right) + 0\left( {0 - 1} \right)} \right|\\ = \dfrac{1}{2}\left| {x\left( { - 1} \right) + 1\left( 2 \right) + 0\left( { - 1} \right)} \right|\\ = \dfrac{1}{2}\left| { - x + 2 + 0} \right|\\ = \dfrac{1}{2}\left| {2 - x} \right|$
It is given that the area of the triangle is $4$ square units.
So, $\dfrac{1}{2}\left| {2 - x} \right| = 4$
Multiply both sides by $2$
$ \Rightarrow \left| {2 - x} \right| = 8$
Modulus sign gives positive value. It means whenever we take out any number from a modulus, it comes out as a positive value. In the case of an expression or an unknown number, it comes out with both positive and negative signs but its value remains positive in an interval.
If the value of $x$ be less than $2$, then $\left( {2 - x} \right)$ will be positive and hence $\left| {2 - x} \right| = \left( {2 - x} \right)$
But if the value of $x$ be greater than $2$, then $\left( {2 - x} \right)$ will be negative and hence $\left| {2 - x} \right| = - \left( {2 - x} \right)$
Here we don’t know whether the value of $\left( {2 - x} \right)$ is positive or not. So, we take $\left| {2 - x} \right| = \pm \left( {2 - x} \right)$
So, $ \pm \left( {2 - x} \right) = 8$
Taking positive sign, we get $2 - x = 8$
Solve this equation.
$ \Rightarrow - x = 8 - 2\\ \Rightarrow - x = 6\\ \Rightarrow x = - 6$
Taking negative sign, we get $ - \left( {2 - x} \right) = 8$
$ \Rightarrow - 2 + x = 8\\ \Rightarrow x = 8 + 2\\ \Rightarrow x = 10$
Finally, we get $x = - 6,10$

Option ‘C’ is correct

Note: You can also solve this question using determinant. In determinant method, area of a triangle having coordinates of the vertices $A\left( {{x_1},{y_1}} \right),B\left( {{x_2},{y_2}} \right),C\left( {{x_3},{y_3}} \right)$ is given by $\dfrac{1}{2}\begin{bmatrix}{{x_1}}&{{y_1}}&1\\{{x_2}}&{{y_2}}&1\\{{x_3}}&{{y_3}}&1 \end{bmatrix}$ and the unit of the area should be taken as square units. Keep in mind that area of a triangle can’t be negative. So, you have to take a modulus sign so that the area can be positive.