Question

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

Answer 1

$\text{Given points are A(-1,2), B(1,5), C(3,4)}$

$\text{Equation of AB is}$ $y-2=\frac{5-2}{1-(-1)}(x+1)$

$\Rightarrow 3x-2y+7=0....\left(i\right)$

$\text{Equation of BC is}$ $y-5=\frac{4-5}{3-1}(x-1)$

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

$\text{Equation of CA is}$ $y-4=\frac{2-4}{-1-3}(x-3)$

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

$\text{Area of}△ABC=\text{Area ABMN + Area NBCK - Area AMKC}$

$={\int }_{-1}^1\frac{3x+7}{2}dx+{\int }_1^3\frac{11-x}{2}dx-{\int }_{-1}^3\frac{x+5}{2}dx$

$=\frac{1}{2}{(\frac{3x^2}{2}+7x)}_{-1}^1+\frac{1}{2}{(11x-\frac{x^2}{2})}_1^3-\frac{1}{2}{(\frac{x^2}{2}+5x)}_{-1}^3$

$=\frac{1}{2}[\frac{3}{2}+7-(\frac{3}{2}-7)]+\frac{1}{2}[33-\frac{9}{2}-(11-\frac{1}{2})]-\frac{1}{2}[\frac{9}{2}+15-(\frac{1}{2}-5)]$

$=\frac{1}{2}(7+7)+\frac{1}{2}(\frac{57}{2}-\frac{21}{2})-\frac{1}{2}(\frac{39}{2}+\frac{9}{2})$

$=\frac{14}{2}+\frac{36}{4}-\frac{48}{4}=7+9-12=4$ $\text{sq.units}$