Question

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

Answer 1

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