Question

find the area of the triangle whose vertices are (-2,1),(3,1),(2,-3)​

Answer 1

$\textbf{Concept:}$

$\text{Area of the triangle formed by the points }(x_1, y_1),\,(x_2, y_2)\text{ and }(x_3, y_3)\text{ is}$

$\boxed{\bf\triangle=\frac{1}{2}[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]}$

$\text{Given points are (-2,1), (3,1), (2,-3) }$

$\text{Now,}$

$\triangle=\frac{1}{2}[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]$

$\triangle=\frac{1}{2}[-2(1+3)+3(-3-1)+2(1-1)]$

$\triangle=\frac{1}{2}[-2(4)+3(-4)+2(0)]$

$\triangle=\frac{1}{2}[-8-12+0]$

$\triangle=\frac{1}{2}[-20]$

$\triangle=-10$

$\text{But area cannot be negative}$

$\therefore\text{Area of the triangle is 10 square units}$

Find more:

Find the area of a quadrilateral formed by the points (1, 0) (0, 1) (- 1, 0) and (0, - 1 )taken in order

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