Question

Using integration, find the area of the region bounded by the triangle whose vertices are $(-1,2),(1,5)\text{and}(3,4).$

Answer 1

First find the equations of the three lines AB,BC,CA.

Equation of AB:
$y-2=\frac{3}{2}(x+1)3x-2y+7=02y=3x+7...\left(i\right)$

Similarly equation of line BC:

$x+2y-11=\mathrm{0...}\left(ii\right)$

and that of line CA:

$x-2y+5=\mathrm{0...}\left(iii\right)$

The area can be found in 2 parts:

Area of triangle ABC=area of triangle ABM+area of triangle AMC

Area of triangle ABM = ${\int }_{-1}^1{(y}_1-y_3)dx={\int }_{-1}^1\{\frac{3x+7}{2}-\frac{x+5}{2}\}dx={\int }_{-1}^1(x+1)dx={[\frac{x^2}{2}+x]}_{-1}^1=\{\frac{1}{2}+1-\frac{1}{2}+1\}=2$

Area of triangle BMC:

${\int }_1^3{(y}_2-y_3)dx={\int }_1^3\{\frac{11-x}{2}-\frac{x+5}{2}\}dx={\int }_1^3(3-x)dx={[3x-\frac{x^2}{2}]}_1^3=\{-\frac{9}{2}+9+\frac{1}{2}-3\}=2$

Hence area of triangle ABC=2+2=4

Thus answer is area of ABC=4