Find the area of the triangle formed by (2, 4),(2, 6),(2+sqrt{3}, 5)

An example on how to find the area of a triangle given the three points vertices (2, 4),(2, 6),(2+sqrt{3}, 5).

$(2,4),(2,6),(2+\sqrt{3},5)\,$

The formula for finding the area of the triangle formed by

$(x_1,y_1),(x_2,y_2),(x_3,y_3)\,$ is

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

Applying the formula,we get

$\frac{1}{2}|[2(6-5)+2(5-4)+(2+\sqrt{3})(4-6)]|\,$

$\frac{1}{2}|2+2-4-2\sqrt{3}|\,$

$\frac{1}{2}|-2\sqrt{3}|\,$

Therefore, the area of the triangle is $\sqrt{3}\,$ .

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