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प्रश्न
Using integration, find the area of region bounded by the triangle whose vertices are (–2, 1), (0, 4) and (2, 3).
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उत्तर
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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