Calculate the area of the triangle ABC with vertices A(2,3), B(1-6), C(0,3)

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We need to find the area of the triangle with the vertexes A(2,3), B(1-6) and C(0,3).

We see that the line AC is perpendicular to the x- axis.

The distance between A and C  is given by sqrt[ ( 2-0)^2 + (3-3)^] = sqrt 2^2 = 2.

The base of...

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We need to find the area of the triangle with the vertexes A(2,3), B(1-6) and C(0,3).

We see that the line AC is perpendicular to the x- axis.

The distance between A and C  is given by sqrt[ ( 2-0)^2 + (3-3)^] = sqrt 2^2 = 2.

The base of the triangle is 2.

The height of the triangle is 3 - (-6) = 3 + 6 = 9

Therefore the area of the triangle is (1/2)*base*height

(1/2)*2*9 = 9 square units.

The required area is 9 square units.

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