Q26 of 45 Page 1

Using integration, find the area of the region bounded by the triangle whose vertices are (-1, 2), (1, 5) and (3, 4).

Plot these points and form a triangle



Write the equations of line AB, BC and AC using two-point form


Equation of line AB,


A = (-1, 2) = (x1, y1) and B = (1, 5) = (x2, y2)




-3(x + 1) = -2(y – 2)


3x + 3 = 2y – 4


2y = 3x + 7


y = 1/2(3x + 7)


Equation of line BC,


B = (1, 5) = (x1, y1) and C = (3, 4) = (x2, y2)




x – 1 = -2(y – 5)


x – 1 = -2y + 10


-2y = x – 11


y = -1/2(x – 11)


Equation of line AC,


A = (-1, 2) = (x1, y1) and C = (3, 4) = (x2, y2)




-2(x + 1) = -4(y – 2)


x + 1 = 2(y – 2)


x + 1 = 2y – 4


2y = x + 5


y = 1/2(x + 5)


Now the area of triangle = area under AB + area under BC – area under AC as shown …(i)



Let us find the area under AB


y = 1/2(3x + 7)


Integrate from -1 to 1










area under AB = 7 unit2 …(a)


the area under BC,


y = -1/2(x – 11)


Integrate from 1 to 3











area under BC = 9 unit2 …(b)


the area under AC,


y = 1/2(x + 5)


Integrate from -1 to 3











area under AC = 7 unit2 …(c)


Using (a), (b) and (c) in (i)


area of triangle = area under AB + area under BC – area under AC


area of triangle = 7 + 9 – 7


area of triangle = 9 unit2


Hence area of triangle is 9 unit2


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