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Question
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 y1= 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.
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