Question

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

Answer 1

Equation of the line passing through points $(2,3)$ and $(3,5)$ is
$\frac{y-3}{x-2}=\frac{5-3}{3-2}\Rightarrow y-3=2x-4\Rightarrow y-2x+1=0\cdots \left(1\right)$

Equation of the line passing through points $(4,4)$ and $(2,3)$ is
$\frac{y-3}{x-2}=\frac{4-3}{4-2}\Rightarrow 2y-6=x-2\Rightarrow 2y-x-4=0\cdots \left(2\right)$

Equation of the line passing through points $(4,4)$ and $(3,5)$ is
$\frac{y-5}{x-3}=\frac{4-5}{4-3}\Rightarrow y-5=3-x\Rightarrow y+x-8=0\cdots \left(3\right)$

Solving $\left(1\right)\&\left(2\right),$ we get the coordinates $x=2,y=3.$
Solving $\left(2\right)\&\left(3\right),$ we get the coordinates $x=4,y=4.$
Solving $\left(3\right)\&\left(1\right),$ we get the coordinates $x=3,y=5.$

Graph of the above equations will be
Required area is
${\int }_2^3(2x-1)dx+{\int }_3^4(8-x)dx-{\int }_2^4\left(\frac{x+4}{2}\right)dx={\left[\frac{(2x-1{)}^2}{2\times 2}\right]}_2^3+{\left[\frac{(8-x{)}^2}{-2}\right]}_3^4-{\left[\frac{(x+4{)}^2}{4}\right]}_2^4=\frac{1}{4}\left[\right(5^2-3^2)-2(4^2-5^2)-(8^2-6^2\left)\right]=\frac{3}{2}\text{sq. units}$