Thank you for reporting, we will resolve it shortly
Q.
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
Application of Integrals
Solution:
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 $ANC$
$=\frac{5}{2} \int\limits_{2}^{4}(x-2) d x+\int\limits_{4}^{6}-(x-9) d x-\frac{3}{4} \int\limits_{2}^{6}(x-2) d x$
$=\frac{5}{2}\left[\frac{(x-2)^{2}}{2}\right]_{2}^{4}-\left[\frac{(x-9)^{2}}{2}\right]_{4}^{6}-\frac{3}{4}\left[\frac{(x-2)}{2}\right]_{2}^{6}$
$=\frac{5}{4}\left[2^{2}-0\right]-\frac{1}{2}\left[(-3)^{2}-(-5)^{2}\right]-\frac{3}{8}\left[(4)^{2}-0\right]$
$=7$ sq. units.
