Question

Using the method of integration, find the area of the triangle $ABC$, coordinates of whose vertices are $A(4,1),B(6,6),C(8,4)$

Answer 1

Area of triangle = Area bounded by line $AB$ from $x=4$ to $6$ + Area bounded by line $BC$ from $x=6$ to $8$ - Area bounded by line $AC$ from $x=4$ to $8$

Area of triangle $={\int }_4^6(\frac{5}{2}x-9)dx+{\int }_6^8(-x+12)dx-{\int }_4^8(\frac{3x}{4}-2)dx$

$=\left|{[\frac{5}{4}{x}^2-9x]}_4^6\right|+\left|{[\frac{-x^2}{2}+12x]}_6^8\right|-\left|{[\frac{3x^2}{8}-2x]}_4^8\right|$

$=|-9+16|+|64-54|-|8+2|=7+10-10=7sq.units$