Question

The area of the triangle with vertices at $\left( { - 4,1} \right),\left( {0,2} \right),\left( {4, - 3} \right)$ is:
A.14
B.16
C.15
D.12

Answer 1

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. A triangle is a two dimensional geometric shape that has 3 sides.

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

Complete step-by-step answer:
We are given three vertices of a triangle.
Let the vertices of the triangle be $A = \left( { - 4,1} \right)$, $B = \left( {0,2} \right)$ and $C = \left( {4, - 3} \right)$.
Hence, we have to find the area of $\Delta ABC$ whose three vertices are given.
Now, substituting $\left( {{x_1},{y_1}} \right) = \left( { - 4,1} \right)$, $\left( {{x_2},{y_2}} \right) = \left( {0,2} \right)$ and $\left( {{x_3},{y_3}} \right) = \left( {4, - 3} \right)$ 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 $\Delta ABC$ $= \left| {\dfrac{1}{2}\left[ { - 4\left( {2 - \left( { - 3} \right)} \right) + 0\left( { - 3 - 1} \right) + 4\left( {1 - 2} \right)} \right]} \right|$
Adding and subtracting the terms in the bracket, we get
$\Rightarrow$ Area of $\Delta ABC$ $= \left| {\dfrac{1}{2}\left[ { - 4\left( {2 + 3} \right) + 0 + 4\left( { - 1} \right)} \right]} \right|$
$\Rightarrow$ Area of $\Delta ABC$ $= \left| {\dfrac{1}{2}\left[ { - 4\left( 5 \right) - 4} \right]} \right|$
Multiplying the terms, we get
$\Rightarrow$ Area of $\Delta ABC$ $= \left| {\dfrac{1}{2}\left( { - 20 - 4} \right)} \right|$
Adding the terms, we get
$\Rightarrow$ Area of $\Delta ABC$ $= \left| {\dfrac{{ - 24}}{2}} \right|$
Dividing $- 24$ by 2, we get
$\Rightarrow$ Area of $\Delta ABC$ $= \left| { - 12} \right|$
Now, we have used the modulus sign because the area of the triangle cannot be negative. So,
Area of \Delta ABC$$$$ = 12 square units
Therefore, the area of the triangle with vertices at $\left( { - 4,1} \right),\left( {0,2} \right),\left( {4, - 3} \right)$ is 12 square units.
Hence, option D is the correct answer.

Note: 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. In other words, 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.