Question

If \[\left( { - 4,0} \right)\]and \[\left( {1, - 1} \right)\] are two vertices of a triangle of area 4 sq. units, then its third vertex lies on
A.\[y = x\]
B.\[5x + y + 12 = 0\]
C.\[x + 5y - 4 = 0\]
D.\[x + 5y + 12 = 0\]

Answer 1

Hint: We will assume the third vertex to be \[(x,y)\] and we will calculate the third vertex using the concept of determinants. Determinant is used to calculate the area of a triangle whose three vertices are given. Here we are given the area and two vertices and so for calculating the area we will work the other way around.

Formula Used:
Let the three vertices of a triangle be \[\left( {{x_1},{y_1}} \right),\left( {{x_2},{y_2}} \right),\left( {{x_3},{y_3}} \right)\], then the area of the triangle using determinant will be given by
\[\dfrac{1}{2}\left| {\begin{array}{*{20}{c}}{{x_1}}&{{y_1}}&1\\{{x_2}}&{{y_2}}&1\\{{x_3}}&{{y_3}}&1\end{array}} \right| = \]area
\[\dfrac{1}{2}\left( {{x_1}\left( {{y_2} - {y_3}} \right) - {y_1}\left( {{x_2} - {x_3}} \right) + \left( {{x_2}{y_3} - {x_3}{y_2}} \right)} \right) = \] |area|

Complete step-by-step answer:

In the question, the area is given to be 4 sq. units. Now, we will plug in the given vertices \[\left( {\left( { - 4,0} \right),\left( {1, - 1} \right)} \right)\], the given area and the assumed vertex \[\left( {x,y} \right)\] into the given formula and solve the question with the following substitutions:
\[{x_1} = x,{\rm{ }}{y_1} = y\]
\[{x_2} = - 4,{\rm{ }}{y_2} = 0\]
\[{x_3} = 1,{\rm{ }}{y_3} = - 1\]
Area = 4
Now, putting the values and evaluating the result:
\[\dfrac{1}{2}\left( {x\left( {0 - \left( { - 1} \right)} \right) - y\left( { - 4 - 1} \right) + \left( {\left( { - 4} \right) \times \left( { - 1} \right) - 0} \right)} \right) = \left| 4 \right|\]
Taking the \[\dfrac{1}{2}\] to the other side and evaluating the expression inside the brackets, we have got,
$\Rightarrow$ \[\left( {x \times 1 - y \times \left( { - 5} \right) + \left( {4 - 0} \right)} \right) = \left| {2 \times 4} \right|\]
$\Rightarrow$ \[\left( {x + 5y + 4} \right) = \left| 8 \right|\]
$\Rightarrow$ \[\left( {x + 5y + 4} \right) = \pm 8\]
Case 1: \[\left( {x + 5y + 4} \right) = 8\]
Taking the \[8\] from the right-hand side to the other side to calculate the final result we have got,
\[x + 5y + 4 - 8 = 0\]
$\Rightarrow$ \[x + 5y - 4 = 0\]
Case 2: \[\left( {x + 5y + 4} \right) = - 8\]
Taking the \[ - 8\] from the right-hand side to the other side to calculate the final result we have got,
\[x + 5y + 4 + 8 = 0\]
$\Rightarrow$ \[x + 5y + 12 = 0\]
Hence, the final result that we got to the question is (C) x + 5y - 4 = 0 and (D) x + 5y + 12 = 0.

Note: In solving the questions like these, it is best to identify the unknown quantity(s), write them down, identify the known quantity(s), note them, then use the appropriate formula or formulas to solve for the values, simplify the complex, complicated expressions, use the basic identities (which must be remembered and be known where to use them) and find the required result. It is to be noted that there are different ways to substitute the unknown value in the formula of the determinant but the end result is always going to be the same.