`(0,0), (3,1), (1,2)` Use calculus to find the area of the triangle with the given vertices.

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Vertices of the triangle are (0,0) , (3,1) and (1,2)

Equation of the line through (0,0) and (3,1) is,

`y-0=((1-0)/(3-0))(x-0)`

`y=x/3`

Equation of the line through (0,0) and(1,2) is,

`y-0=((2-0)/(1-0))(x-0)`

`y=2x`

Equation of the line through (3,1) and (1,2) is

`y-1=((2-1)/(1-3))(x-3)`

`y-1=(x-3)/-2`

`y=-1/2x+3/2+1`

`y=-1/2x+5/2`

Refer to the attached image .Graph of the lines is plotted. 

Area of the triangle A `=int_0^1(2x-x/3)dx+int_1^3((-x/2+5/2)-x/3)dx`

`A=int_0^1(5x)/3+int_1^3((-3x-2x)/6+5/2)dx`

`A=[5/3(x^2/2)]_0^1+int_1^3((-5x)/6+5/2)dx`

`A=[(5x^2)/6]_0^1+[-5/6(x^2/2)+5/2x]_1^3`

`A=[5/6(1)^2-5/6(0)^2]+[-5/12(3)^2+5/2(3)]-[-5/6(1)^2+5/2(1)]`

`A=5/6+(-15/4+15/2)-(-5/12+5/2)`

`A=5/6+(-15+30)/4-(-5+30)/12`

`A=5/6+15/4-25/12`

`A=30/12`

`A=5/2`

 

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