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)
\(=∫^4_2\frac{5}{2}(x-2)dx+∫^6_4(-x+9)dx-∫^6_2\frac{3}{4}(x-2)dx\)
\(=\frac{5}{2}\bigg[\frac{x^2}{2}-2x\bigg]^4_2+\bigg[\frac{-x^2}{2}+9x\bigg]^6_4-\frac{3}{4}\bigg[\frac{x^2}{2}-2x\bigg]^6_2\)
\(=\frac{5}{2}[8-8-2+4]+[-18+54+8-36]-\frac{3}{4}[18-12-2+4]\)
\(=5+8-\frac{3}{4}(8)\)
\(=13-6\)
\(=7units\)