Question

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

• A. 2
• B. 4
• C. 7
• D. 8

Answer 1

Answer: (C)

Equation of the line $AB$ is $y=\frac{5}{2}(x-2)$
Equation of the line $BC$ is $y=-x+9$

Equation of the line $CA$ is $y=\frac{3}{4}(x-2)$
Required area $=$ area of the region bounded by $\Delta ABC$ $=$ area of the region $AMB+$ Area of region $BMNC-$ area of the region $ANC$
$=\frac{5}{2}\underset{2}{\overset{4}{\int }}(x-2)dx+\underset{4}{\overset{6}{\int }}-(x-9)dx-\frac{3}{4}\underset{2}{\overset{6}{\int }}(x-2)dx$
$=\frac{5}{2}{\left[\frac{(x-2{)}^2}{2}\right]}_2^4-{\left[\frac{(x-9{)}^2}{2}\right]}_4^6-\frac{3}{4}{\left[\frac{(x-2)}{2}\right]}_2^6$
$=\frac{5}{4}\left[2^2-0\right]-\frac{1}{2}\left[(-3{)}^2-(-5{)}^2\right]-\frac{3}{8}\left[(4{)}^2-0\right]$
$=7$ sq. units.