Question

Using the method of integration find the area of the triangle $ABC$, coordinates of whose vertices are $A(2,0),B(4,5)$ and $C(6,3)$

Answer 1

The vertices of $\Delta ABC$ are $A(2,0),B(4,5)$, and $C(6,3)$.

Equation of line segment $AB$ is $y-0=\frac{5-0}{4-2}(x-2)$
$2y=5x-10$
$y=\frac{5}{2}(x-2)..........\left(1\right)$

Equation of line segment $BC$ is $y-5=\frac{3-5}{6-4}(x-4)$
$2y-10=-2x+8$
$2y=-2x+18$
$y=-x+9.........\left(2\right)$

Equation of line segment $CA$ is $y-3=\frac{0-3}{2-6}(x-6)$
$-4y+12=-3x+18$
$4y=3x-6$
$y=\frac{3}{4}(x-2)...........\left(3\right)$

$\therefore Area\left(\Delta ABC\right)=Area\left(ABLA\right)+Area\left(BLMCB\right)-Area\left(ACMA\right)$

$={\int }_2^4\frac{5}{2}(x-2)dx+{\int }_4^6(-x+9)dx-{\int }_2^6\frac{3}{4}(x-2)dx$

$=\frac{5}{2}{[\frac{x^2}{2}-2x]}_2^4+{[\frac{-x^2}{2}+9x]}_4^6-\frac{3}{4}{[\frac{x^2}{2}-2x]}_2^6$

$=\frac{5}{2}[8-8-2+4]+[-18+54+8-36]-\frac{3}{4}[18-12-2+4]$

$=5+8-\frac{3}{4}\left(8\right)$

$=13-6=7$ sq. units