Using integration, find the area of the triangle whose vertices are (2,3),(3,5), and (4,4).
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Solution
Equation of the line passing through points (2,3) and (3,5) is y−3x−2=5−33−2⇒y−3=2x−4⇒y−2x+1=0⋯(1)
Equation of the line passing through points (4,4) and (2,3) is y−3x−2=4−34−2⇒2y−6=x−2⇒2y−x−4=0⋯(2)
Equation of the line passing through points (4,4) and (3,5) is y−5x−3=4−54−3⇒y−5=3−x⇒y+x−8=0⋯(3)
Solving (1)&(2), we get the coordinates x=2,y=3.
Solving (2)&(3), we get the coordinates x=4,y=4.
Solving (3)&(1), we get the coordinates x=3,y=5.
Graph of the above equations will be
Required area is ∫32(2x−1)dx+∫43(8−x)dx−∫42(x+42)dx=[(2x−1)22×2]32+[(8−x)2−2]43−[(x+4)24]42=14[(52−32)−2(42−52)−(82−62)]=32sq. units