Question

The coordinates of the three vertices of the triangle are $(8, - 5),( - 2, - 7)$and$(5,1)$, find the area of the triangle.

Answer 1

Hint: Area is called the space occupied by the figure. The unit of area is squares, it is termed as square centimeters, square inches, square feet etc. In this problem, we have to find the area of the triangle whose vertices are given i.e. $(8, - 5),( - 2, - 7)$ and $(5,1)$.

Formula Used:
Area of triangle$ = \dfrac{1}{2}\left| {{x_1}({y_2} - {y_3}) + {x_2}({y_3} - {y_1}) + {x_3}({y_1} - {y_2})} \right|$
Where, $({x_1},{y_1}),({x_2},{y_2})$ and $({x_3},{y_3})$ are the coordinates of the vertices of the triangle.

Complete step by step answer:
Area of triangle $ = \dfrac{1}{2} \times b \times h$, for the right angle triangle but here the coordinates of the vertices of the triangle are given and to find the area of the triangle, there is a formula i.e.,
Area of triangle$ = \dfrac{1}{2}\left| {{x_1}({y_2} - {y_3}) + {x_2}({y_3} - {y_1}) + {x_3}({y_1} - {y_2})} \right|$
Where, $({x_1},{y_1}),({x_2},{y_2})$ and $({x_3},{y_3})$are the coordinates of the vertices of the triangle.
Let us assume, Area of triangle$ = A$,${x_1} = 8,{y_1} = - 5,{x_2} = - 2,{y_2} = - 7,{x_3} = 5,{y_3} = 1$. Now, let us substitute these values into the above formula.
$ \Rightarrow A = \dfrac{1}{2}\left| {8( - 7 - 1) + ( - 2)(1 - ( - 5)) + 5( - 5 - ( - 7))} \right|$
On further solving, we get,
$
   \Rightarrow A = \dfrac{1}{2}\left| {8( - 8) - 2(1 + 5) + 5( - 5 + 7)} \right| \\
   \Rightarrow A = \dfrac{1}{2}\left| { - 64 - 2(6) + 5(2)} \right| \\
   \Rightarrow A = \dfrac{1}{2}\left| { - 64 - 12 + 10} \right| \\
   \Rightarrow A = \dfrac{1}{2}\left| { - 66} \right| \\
   \Rightarrow A = \left| { - 33} \right| \\
   \Rightarrow A = 33 \\
 $
Hence, the area of the triangle from the coordinates of the vertices is $33c{m^2}$.

Note: In this formula, we have used a modulus sign, the modulus is also known as the absolute value of a real number. The sign of the modulus is $\left| {} \right|$. If a positive number is placed inside this modulus sign then it results in a positive number. For example,$\left| 3 \right| = 3$. But if a negative number is placed inside the modulus sign then it results in a positive number. For example, $\left| { - 2} \right| = 2$. Thus, the absolute value is always either positive or zero.