Question

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

Answer 1

Given triangle $ABC$, coordinates of whose vertices are $A(4,1),B(6,6)$ and $C(8,4)$.

Equation of $AB$ is given by

$y-1=\frac{6-1}{6-4}(x-4)ory=\frac{5}{2}x-9$

Equation of $BC$ is given by

$y-6=\frac{4-6}{8-6}(x-6)$ or $y=-x+12$

Equation of $AC$ is given by

$y-1=\frac{4-1}{8-4}(x-4)$ or $y=\frac{3}{4}x-2$

$\therefore$ Area of $△ABC$ = area of trap. $DABE+$ area of trap. $EBCF$ - area of trap. $DACF$

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

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

=$\frac{5}{4}(36-16)-9(6-4)-\frac{1}{2}(64-36)+12(8-6)-\frac{3}{8}(64-16)+2(8-4)$

=$\frac{5}{4}\times 20-18-\frac{28}{2}+24-\frac{3}{8}\times 48+8$

$=25-18-14+24-18+8=7$ sq units