
Calculate the area of a triangle with vertices $\left( {1,1} \right),\left( {3,1} \right)$ and $\left( {5,7} \right)$.
A. 6
B. 7
C. 9
D. 10
Answer
464.4k+ views
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|$
Complete step-by-step answer:
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.
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|$
Complete step-by-step answer:
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.
Recently Updated Pages
Basicity of sulphurous acid and sulphuric acid are

Master Class 12 Economics: Engaging Questions & Answers for Success

Master Class 12 Maths: Engaging Questions & Answers for Success

Master Class 12 Biology: Engaging Questions & Answers for Success

Master Class 12 Physics: Engaging Questions & Answers for Success

Master Class 4 Maths: Engaging Questions & Answers for Success

Trending doubts
In Indian rupees 1 trillion is equal to how many c class 8 maths CBSE

List some examples of Rabi and Kharif crops class 8 biology CBSE

How many ounces are in 500 mL class 8 maths CBSE

What is the feminine gender of a stag class 8 english CBSE

Give me the opposite gender of Duck class 8 english CBSE

Summary of the poem Where the Mind is Without Fear class 8 english CBSE
