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 correct answer is:$7units$
The vertices of ∆ABC are A(2,0),B(4,5),and C(6,3).

Equation of line segment AB is
$y-0=\frac{5-0}{4-2}(x-2)$
$2y=5x-10$
$y=\frac{5}{2}(x-2)...(1)$
Equation of line segment BC is
$y-5=\frac{3-5}{6-4}(x-4)$
$2y-10=-2x+8$
$2y=-2x+18$
$y=-x+9...(2)$
Equation of line segment CA is
$y-3=\frac{0-3}{2-6}(x-6)$
$-4y+12=-3x+18$
$4y=3x-6$
$y=\frac{3}{4}(x-2)...(3)$
Area(ΔABC)=Area(ABLA)+Area(BLMCB)-Area(ACMA)
$=∫^4_2\frac{5}{2}(x-2)dx+∫^6_4(-x+9)dx-∫^6_2\frac{3}{4}(x-2)dx$
$=\frac{5}{2}\bigg[\frac{x^2}{2}-2x\bigg]^4_2+\bigg[\frac{-x^2}{2}+9x\bigg]^6_4-\frac{3}{4}\bigg[\frac{x^2}{2}-2x\bigg]^6_2$
$=\frac{5}{2}[8-8-2+4]+[-18+54+8-36]-\frac{3}{4}[18-12-2+4]$
$=5+8-\frac{3}{4}(8)$
$=13-6$
$=7units$