Question

Using integration, find the area of the triangle ABC. Coordinates of whose vertices are A(2, 0), B(4, 5) and C(6 ; 3).

Answer 1

Given,

Vertices of triangle$\[A\left( {2,0} \right),B\left( {4,5} \right),C\left( {6,3} \right)\]$

Solution,

Know that the area of triangle is $\[\frac{1}{2}\left\{ {{x_1}\left( {{y_2} - {y_3}} \right) + {x_2}\left( {{y_3} - {y_1}} \right) + {x_3}\left( {{y_1} - {y_2}} \right)} \right\}\]$ when vertices are $\[\left( {{x_1},{y_1}} \right),\left( {{x_2},{y_2}} \right),\left( {{x_3},{y_3}} \right)\]$

Calculate the area of the triangle.

$\[\begin{array}{l} = \frac{1}{2}\left\{ {{x_1}\left( {{y_2} - {y_3}} \right) + {x_2}\left( {{y_3} - {y_1}} \right) + {x_3}\left( {{y_1} - {y_2}} \right)} \right\}\\ = \frac{1}{2}\left\{ {2\left( {5 - 3} \right) + 4\left( {3 - 0} \right) + 6\left( {0 - 5} \right)} \right\}\\ = \frac{1}{2}\left\{ {2 \times 2 + 4 \times 3 + 6 \times \left( { - 5} \right)} \right\}\\ = \frac{1}{2}\left( {4 + 12 - 30} \right)\\ = - 7\,uni{t^2}\end{array}\]$

Area can not be negative so neglect the negative sign.

Hence the area of triangle is $\[7\,uni{t^2}\]$.