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Question: How to find the area of a triangle whose vertices are: \(\left( {3, - 1} \right)\), \(\left( {2,7} \...

How to find the area of a triangle whose vertices are: (3,−1)\left( {3, - 1} \right), (2,7)\left( {2,7} \right) and (−5,6)\left( { - 5,6} \right)?

Explanation

Solution

In the given question, we are required to find the area of the triangle whose coordinates of vertices are given to us in the problem. So, we substitute the values of coordinates into the formula for area of the triangle as 12∣x1(y2−y3)+x2(y3−y1)+x3(y1−y2)∣\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| where (x1,y1)\left( {{x_1},{y_1}} \right), (x2,y2)\left( {{x_2},{y_2}} \right) and (x3,y3)\left( {{x_3},{y_3}} \right) are coordinates of the vertices of triangle.

Complete step by step answer:
The points given to us in the question are: (3,−1)\left( {3, - 1} \right), (2,7)\left( {2,7} \right) and (−5,6)\left( { - 5,6} \right)
Now, we know that the formula for the area of triangle ABC with coordinates of vertices as (x1,y1)\left( {{x_1},{y_1}} \right), (x2,y2)\left( {{x_2},{y_2}} \right) and (x3,y3)\left( {{x_3},{y_3}} \right) is 12∣x1(y2−y3)+x2(y3−y1)+x3(y1−y2)∣\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|.

So, we have, (x1,y1)=(3,−1)\left( {{x_1},{y_1}} \right) = \left( {3, - 1} \right), (x2,y2)=(2,7)\left( {{x_2},{y_2}} \right) = \left( {2,7} \right) and (x3,y3)=(−5,6)\left( {{x_3},{y_3}} \right) = \left( { - 5,6} \right).
Substituting the values of coordinates of the three points, we get,
⇒12∣3(7−6)+2(6−(−1))−5(−1−7)∣\Rightarrow \dfrac{1}{2}\left| {3\left( {7 - 6} \right) + 2\left( {6 - \left( { - 1} \right)} \right) - 5\left( { - 1 - 7} \right)} \right|
Simplifying the expression,
⇒12∣3(1)+2(6+1)−5(−8)∣\Rightarrow \dfrac{1}{2}\left| {3\left( 1 \right) + 2\left( {6 + 1} \right) - 5\left( { - 8} \right)} \right|

Opening the brackets, we get,
⇒12∣3+14+40∣\Rightarrow \dfrac{1}{2}\left| {3 + 14 + 40} \right|
Simplifying the expression we get,
⇒12∣57∣\Rightarrow \dfrac{1}{2}\left| {57} \right|
So, we know that the modulus of any positive number is the number itself. So, we get,
⇒572\Rightarrow \dfrac{{57}}{2}
Expressing in decimal representation, we get,
⇒28.5\Rightarrow 28.5

So, the area of the triangle whose vertices are given as (3,−1)\left( {3, - 1} \right), (2,7)\left( {2,7} \right) and (−5,6)\left( { - 5,6} \right) is 28.528.5 square units.

Note: Modulus function always returns the magnitude of the number which is always positive. Hence, even a negative number in a modulus function yields a positive value. We can take the coordinates of the vertices of the triangle in any cyclic order, we will arrive at the same answer as the sign will be adjusted due to the modulus function.