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)
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⇒ -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)
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⇒ 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)
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⇒ -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
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⇒ area under AB = 7 unit2 …(a)
the area under BC,
⇒ y = -1/2(x – 11)
Integrate from 1 to 3
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⇒ area under BC = 9 unit2 …(b)
the area under AC,
⇒ y = 1/2(x + 5)
Integrate from -1 to 3
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⇒ 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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