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Question
Using integration, find the area of the region bounded by the triangle whose vertices are (−1, 2), (1, 5) and (3, 4).
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Solution
Let ABC be the required triangle with vertices A (−1, 2), B (1, 5) and C (3, 4).
Now, the equation of the side AB is
`y−2=(5−2)/(1−(−1))(x−(−1))`
`⇒ 3x − 2y + 7 = 0 ... (i)`
The equation of BC is
`y−5=(4−5)/(3−1)(x−1)`
`⇒ x + 2y − 11 = 0 ... (ii)`
The equation of AC is
`y−2=(4−2)/(3−(−1))(x−(−1))`
`⇒ x − 2y + 5 = 0 ... (iii)`
Now, area of ΔABC `= ∫_(−1)^1yABdx+∫_1^3yBCdx−∫_(−1)^3yACdx`
`=∫_(-1)^1(3x+7)/2dx + ∫_1^3(11−x)/2dx − ∫_1^3(x+5)/2dx`
`=12[((3x^2)/2+7x)_(-1)^1+(11x−x^2/2)_1^3−(x^2/2+5x)_(-1)^3]`
`=1/2[(3/2+7−3/2+7)+(33−9/2−11+12)−(9/2+15−1/2+5)]`
`= 1/2 × 8 = 4 sq. units`
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