Question

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

Answer 1

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`