Question

Find the area of the triangle formed by the points $ \left( {8, - 5} \right) $ , $ \left( { - 2, - 7} \right) $ and $ \left( {5,1} \right) $ .

Answer 1

Hint: In this question, we need to determine the area of the triangle such that the triangle is formed by the points $ \left( {8, - 5} \right) $ , $ \left( { - 2, - 7} \right) $ and $ \left( {5,1} \right) $ .For this we will use the distance formula to determine the length of the sides of the triangle and then, apply Heron's formula to evaluate the area of the triangle.

Complete step-by-step answer:
Let the points of the triangle be A, B, C .Where, $ A = \left( {8, - 5} \right) $ , $ B = \left( { - 2, - 7} \right) $ and $ C = \left( {5,1} \right) $ .

Now, we will find the sides of the triangle AB, BC, AC using the distance formula.
First, let us find AC using the distance formula,
 $ d = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} $
Where, $ \left( {{x_1},{y_1}} \right) = \left( {8, - 5} \right) $ and $ \left( {{x_2},{y_2}} \right) = \left( { - 2, - 7} \right) $
Now, substituting the values in the distance formula,
 $
  AB = \sqrt {{{\left( { - 2 - 8} \right)}^2} + {{\left( { - 7 + 5} \right)}^2}} \\
   \Rightarrow AB = \sqrt {{{\left( { - 10} \right)}^2} + {{\left( { - 2} \right)}^2}} \\
   \Rightarrow AB = \sqrt {100 + 4} \\
   \Rightarrow AB = \sqrt {104} \\
   \Rightarrow AB = 10.19 \;
  $
Now, to find BC,
Let us consider, $ \left( {{x_1},{y_1}} \right) = \left( { - 2, - 7} \right) $ and $ \left( {{x_2},{y_2}} \right) = \left( {5,1} \right) $
Now, substituting the values in the distance formula,
 $
  BC = \sqrt {{{\left( {5 + 2} \right)}^2} + {{\left( {1 + 7} \right)}^2}} \\
   \Rightarrow BC = \sqrt {{{\left( 7 \right)}^2} + {{\left( 8 \right)}^2}} \\
   \Rightarrow BC = \sqrt {49 + 64} \\
   \Rightarrow BC = \sqrt {113} \\
   \Rightarrow BC = 10.63 \;
  $
Now, to find AC,
Let us consider, $ \left( {{x_1},{y_1}} \right) = \left( {8, - 5} \right) $ and $ \left( {{x_2},{y_2}} \right) = \left( {5,1} \right) $
Now, substituting the values in the distance formula,
 $
  AC = \sqrt {{{\left( {5 - 8} \right)}^2} + {{\left( {1 + 5} \right)}^2}} \\
   \Rightarrow AC = \sqrt {{{\left( { - 3} \right)}^2} + {{\left( 6 \right)}^2}} \\
   \Rightarrow AC = \sqrt {9 + 36} \\
   \Rightarrow AC = \sqrt {45} \\
   \Rightarrow AC = 6.70 \;
  $
Therefore, the sides of the triangle are, $ AB = 10.19 $ , $ BC = 10.63 $ and $ AC = 6.70 $
Now, to find the area of the triangle, let us use the Heron's formula,
 $ Area = \sqrt {s\left( {s - a} \right)\left( {s - b} \right)\left( {s - c} \right)} $
Where, $ s =\dfrac{{a + b + c}}{2} $
Here $ a,b,c $ are the side lengths of the triangle.
Let\[a = AB\], $ b = BC $ and $ c = AC $
Now, substituting the values in $ s =\dfrac{{a + b + c}}{2} $ ,
 $
  s =\dfrac{{10.19 + 10.63 + 6.70}}{2} \\
   \Rightarrow s = 13.76 \;
  $
Now applying the value of s in the Heron's formula,
 $
  Area = \sqrt {13.76\left( {13.76 - 10.19} \right)\left( {13.76 - 10.63} \right)\left( {13.76 - 6.70} \right)} \\
   \Rightarrow Area = 32.947 \;
  $
Therefore, the area of the given triangle is $ 32.947sq.uts $ . $ 32.947\;sq.units $
So, the correct answer is “ $ 32.947\;sq.units $ ”.

Note: It is important to note here that the Heron's formula can only be used to find the area of the triangle of which the length of each side has been given. As the points of the triangle are given, we can also use the formula $ Area =\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) $ . It must be taken care of by the candidates to use the appropriate coordinates while evaluating the distance between the points.
Alternatively,
\[
  \Delta =\dfrac{1}{2}\left| {\left( {\begin{array}{\times{20}{c}}
  {{x_1}}&{{y_1}}&1 \\
  {{x_2}}&{{y_2}}&1 \\
  {{x_3}}&{{y_3}}&1
\end{array}} \right)} \right| \\
   \Rightarrow \Delta =\dfrac{1}{2}\left| {\left( {\begin{array}{\times{20}{c}}
  8&{ - 5}&1 \\
  { - 2}&{ - 7}&1 \\
  5&1&1
\end{array}} \right)} \right| \\
   \Rightarrow \Delta =\dfrac{1}{2}\left[ {8\left( { - 7\times1 - 1} \right) + 5( - 2 - 5) + 1( - 2\times1 + 7\times5)} \right] \\
   \Rightarrow \Delta =\dfrac{1}{2}\left[ {8\left( { - 8} \right) + 5( - 7) + 1(33)} \right] \\
   \Rightarrow \Delta =\dfrac{1}{2}\left[ { - 64 - 35 + 33} \right] \\
   \Rightarrow \Delta =\dfrac{1}{2}\left[ { - 66} \right] \\
   \Rightarrow \Delta = 33 \;
 \]