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Calculate the area of a triangle with vertices $\left( {1,1} \right),\left( {3,1} \right)$ and $\left( {5,7} \right)$.A. 6B. 7C. 9D. 10

Last updated date: 22nd Feb 2024
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Hint: Here, will use the formula of the area of a triangle and substitute the given vertices in that formula to find the required area. A triangle is a two-dimensional figure which has three sides and three vertices.

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

According to the question,
We are given the three vertices of a triangle.
Let the vertices of the triangle be $A = \left( {1,1} \right)$, $B = \left( {3,1} \right)$ and $C = \left( {5,7} \right)$.
Hence, we have to find the area of $\vartriangle ABC$ whose three vertices are given.
Now, we will use the formula:
Area of a triangle $= \left| {\dfrac{1}{2}\left[ {{x_1}\left( {{y_2} - {y_3}} \right) + {x_2}\left( {{y_3} - {y_1}} \right) + {x_3}\left( {{y_1} - {y_2}} \right)} \right]} \right|$………………….$\left( 1 \right)$
Now, substituting $\left( {{x_1},{y_1}} \right) = \left( {1,1} \right)$, $\left( {{x_2},{y_2}} \right) = \left( {3,1} \right)$ and $\left( {{x_3},{y_3}} \right) = \left( {5,7} \right)$, in equation $\left( 1 \right)$, we get
Area of $\vartriangle ABC = \left| {\dfrac{1}{2}\left[ {1\left( {1 - 7} \right) + 3\left( {7 - 1} \right) + 5\left( {1 - 1} \right)} \right]} \right|$
$\Rightarrow$ Area of $\vartriangle ABC = \left| {\dfrac{1}{2}\left[ {1\left( { - 6} \right) + 3\left( 6 \right) + 5\left( 0 \right)} \right]} \right|$
Solving this further, we get,
$\Rightarrow$ Area of $\vartriangle ABC = \left| {\dfrac{1}{2}\left[ { - 6 + 18 + 0} \right]} \right| = \left| {\dfrac{1}{2}\left( {12} \right)} \right| = \left| 6 \right|$
Now, we have used the modulus sign because the area of a triangle cannot be negative.
Hence, Area of $\vartriangle ABC = 6$ square units
Therefore, the area of the triangle with vertices $\left( {1,1} \right),\left( {3,1} \right)$and $\left( {5,7} \right)$ is 6 square units.

Hence, option A is the correct answer.

Note:
We can also find the area of the triangle using the help of determinants.
We will use the formula:
Area of triangle
$= \left| {\dfrac{1}{2}\left| \begin{gathered} {x_1}{\text{ }}{y_1}{\text{ }}1 \\ {x_2}{\text{ }}{y_2}{\text{ }}1 \\ {x_3}{\text{ }}{y_3}{\text{ }}1 \\ \end{gathered} \right|} \right|$
Now, Substituting $\left( {{x_1},{y_1}} \right) = \left( {1,1} \right)$, $\left( {{x_2},{y_2}} \right) = \left( {3,1} \right)$ and $\left( {{x_3},{y_3}} \right) = \left( {5,7} \right)$ we get,
Area of $\vartriangle ABC = \left| {\dfrac{1}{2}\left| {\begin{array}{*{20}{l}} 1&1&1 \\ 3&1&1 \\ 5&7&1 \end{array}} \right|} \right|$
Now, solving the determinant,
$\Rightarrow ar\vartriangle ABC = \left| {\dfrac{1}{2}\left[ {1\left( {1 - 7} \right) - 1\left( {3 - 5} \right) + 1\left( {21 - 5} \right)} \right]} \right|$
$\Rightarrow ar\vartriangle ABC = \left| {\dfrac{1}{2}\left[ { - 6 + 2 + 16} \right]} \right|$
Solving further,
$\Rightarrow ar\vartriangle ABC = \left| {\dfrac{1}{2} \times 12} \right| = \left| 6 \right|$
Therefore, area of $\vartriangle ABC = 6$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.