Question

Using definite integration, find area of the triangle with vertices at A(1,1),B(3,3)A(1,1),B(3,3).

Answer 1

$e{q}^{n}ofAB:$

$y-1=\frac{2}{2}\left(x-1\right)$

$y-1=x-1$

$y=x$

$e{q}^{n}ofBC:$

$y-3=\frac{-2}{1}\left(x-3\right)$

$y-3=-2x+6$

$2x+y=9$

$Area={\int }_1^{3}xdx+{\int }_3^4\left(9-2x\right)dx-\underset{1}{\overset{4}{\int }}1dx$

$={\left(\frac{x^2}{2}\right)}_1^3+{\left(9x-x^2\right)}_3^4-3$

$=\frac{9}{2}-\frac{1}{2}+\left[\left(36-16\right)-\left(27-9\right)\right]-3$

$=\frac{8}{2}+\left(20-18\right)-3$

$=4+2-3$

$=3$.