Using integration, find the area of triangle ABC, whose vertices are A(2, 5), B(4, 7) and C(6, 2).

#### Solution

Vertices of the given triangle are A(2,5), B(4,7), and C(6,2).

Equation of AB

`y - 5 = (7-5)/(4-2) (x -2)`

⇒ `y - 5 = x -2`

⇒ `y = x + 3`

Let's say y_{1}= x+3

Equation of BC:

`y -7= (2-7)/(6-4)(x-4)`

⇒ `y = (-5)/(2) (x-4) +7=(-5)/(2) x+17`

Let's say `y_2 = -(5)/(2)x+17`

Equation of AC:

`y -5 = (2-5)/(6-2) (x-2)`

⇒ `y = (-3)/(4)(x-2)+5 = (-3)/(4)x+13/2`

Let's say `y_3 = (-3)/(4)x+13/2`

`"ar" (Δ"ABC") = int_2^4 y_1 dx + int_4^6 y_2 dx - int_2^6 y_3 dx`

= `int_2^4 (x+3) dx + int_4^6 ((-5)/2 x+17) dx -int_2^6 (-3)/4 x+13/2)dx`

= `[x^2/2 + 3x]_2^4 + [(-5x^2)/4 + 17x]_4^6 - [(-3x^2)/8 + (13x)/2]_2^6`

= `[16/2 + 12 - 4/2 -6]+[(-180)/4 + 102+80/4-68]-[(-108)/8+78/2+12/8-26/2]`

= 12 + 9 - 14

= 7 sq units.