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

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justaguide's profile pic

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

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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giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

The area of the triangle ABC coulde be calculated using the formula:

S = (1/2)*|det M|

              |xA    yA    1|

det M =  |xB    yB    1|

              |xC    yC    1|

det M = xA*yB + xB*yC + xC*yA - xCyB - xA*yC - xB*yA

We'll substitute the coordinates of the vertices A,B,C:

det M = 2*(-6) + 1*3 + 0*3 - 0*(-6) - 2*3 - 1*3

det M = -12 + 3 - 6 - 3

det M = -18

The area of the triangle is:

S = (1/2)*|-18|

S = 9 square units.

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