Question

Using integration find area of triangle ABC whose vertices have coordinates A(2,5) B(4,7) C(6,2)

Answer 1

the area of triangle of vertices A(2,5) B(4,7) C(6,2) using the formula 1/2|X1(Y2-Y3)+X2(Y3-Y1)+X3(Y1-Y2)| where A = (X1,Y1) B=(X2,Y2 C=(X3,Y3) is 7. the are of triangle is 7units.

Answer 2

The area of the triangle ABC whose vertices are A (2, 5), B (4, 7) and C (6, 2) is 7 sq. units.

Step-by-step explanation:

[ Refer to the given attachment for the diagram. ]

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The equation of the side AB is

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

or, (x - 4)/2 = (y - 7)/2

or, x - 4 = y - 7

or, x - y + 3 = 0 ..... (1)

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The equation of the side BC is

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

or, (x - 6)/2 = (y - 2)/(- 5)

or, - 5x + 30 = 2y - 4

or, 5x + 2y = 34 ..... (2)

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The equation of the side CA is

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

or, (x - 2)/(- 4) = (y - 5)/3

or, 3x - 6 = - 4y + 20

or, 3x + 4y = 26 ..... (3)

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Therefore, the area of the triangle ABC is

= (the area made by the side AB on the x-axis - the area made by the side CA on the x-axis in the region of AB) + (the area made by the side BC on the x-axis - the area made by the side CA on the x-axis in the region of BC)

= ∫₂⁴ [ (x + 3) - (26 - 3x)/4 ] dx + ∫₄⁶ [ (34 - 5x)/2 - (26 - 3x)/4 ] dx

= [ x²/2 + 3x - 13x/2 + 3x²/8 ]₂⁴ + [ 17x - 5x²/4 - 13x/2 + 3x²/8 ]₄⁶

= [ 7x²/8 - 7x/2 ]₂⁴ + [ 21x/2 - 7x²/8 ]₄⁶

= {(112/8 - 28/2) - (28/8 - 14/2)} + {(126/2 - 252/8) - (84/2 - 112/8)} sq. units

= {(112 - 112)/8 - (28 - 56)/8} + {(504 - 252)/8 - (336 - 112)/8} sq. units

= {0 - (- 28/8)} + (252/8 - 224/8) sq. units

= 28/8 + (252 - 224)/8 sq. units

= 28/8 + 28/8 sq. units

= (28 + 28)/8 sq. units

= 56/8 sq. units

= 7 sq. units

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NOTE:

We can try solving the problem in another way where we add the regions formed by the sides AB, BC on the x-axis and subtract the region formed by the side CA on the x-axis. Find the limits of the integration carefully from diagram.