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

*print*Print*list*Cite

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.**

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.**