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Question
Using integration, find the area of region bounded by the triangle whose vertices are (–2, 1), (0, 4) and (2, 3).
Solution
Let the vertices of the triangle be A(–2, 1), B(0, 4) and C(2, 3).
Equation of line segment AB is
`(y - 1) = (4-1)/(0+2) (x + 2)`
`=>y = (3(x + 2))/2 + 1`
`=> y = (3x + 6 + 2)/2`
`=> y = (3x + 8)/2`
Equation of line segment BC is
`(y - 4) = (3-4)/(2 - 0) (x - 0)`
`=> y = -1/2x + 4`
Equation of line segment AC is
`(y - 3) = (1-3)/(-2-2) (x - 2)`
`=> y = 1/2(x - 2) + 3`
`=> y = x/2 - 1 + 3`
`=> y = x/2 + 2`
The graph of the given equations is shown below:
AL, BO and CM are drawn perpendicular to x-axis.
It can be observed in the following figure that,
Area (ΔACB) = Area (ABOL) + Area (BOMCB) – Area (ALMCA) ….. (1)
Required area = Area (ΔACB) = Area (ABOL) + Area (BOMCB) – Area (ALMCA)
= 5 + 7 − 8
= 4 sq units
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