Using the method of integration,find the area | vedclass

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(A) The vertices of $\Delta ABC$ are $A(2,0)$,$B(4,5)$,and $C(6,3)$.Equation of line segment $AB$ passing through $(2,0)$ and $(4,5)$ is:$y - 0 = \fra

Vedclass · Accepted Answer

$7 	ext{ sq. units}$

Vedclass · Answer

(A) The vertices of $\Delta ABC$ are $A(2,0)$,$B(4,5)$,and $C(6,3)$.Equation of line segment $AB$ passing through $(2,0)$ and $(4,5)$ is:$y - 0 = \frac{5-0}{4-2}(x-2) \implies y = \frac{5}{2}(x-2) \quad \dots(1)$Equation of line segment $BC$ passing through $(4,5)$ and $(6,3)$ is:$y - 5 = \frac{3-5}{6-4}(x-4) \implies y - 5 = -1(x-4) \implies y = -x + 9 \quad \dots(2)$Equation of line segment $AC$ passing through $(2,0)$ and $(6,3)$ is:$y - 0 = \frac{3-0}{6-2}(x-2) \implies y = \frac{3}{4}(x-2) \quad \dots(3)$Area of $\Delta ABC = \int_{2}^{4} y_{AB} \, dx + \int_{4}^{6} y_{BC} \, dx - \int_{2}^{6} y_{AC} \, dx$$= \int_{2}^{4} \frac{5}{2}(x-2) \, dx + \int_{4}^{6} (-x+9) \, dx - \int_{2}^{6} \frac{3}{4}(x-2) \, dx$$= \frac{5}{2} \left[ \frac{(x-2)^2}{2} ight]_{2}^{4} + \left[ -\frac{x^2}{2} + 9x ight]_{4}^{6} - \frac{3}{4} \left[ \frac{(x-2)^2}{2} ight]_{2}^{6}$$= \frac{5}{2} \left( \frac{4}{2} - 0 ight) + \left( (-18 + 54) - (-8 + 36) ight) - \frac{3}{4} \left( \frac{16}{2} - 0 ight)$$= 5 + (36 - 28) - \frac{3}{4}(8)$$= 5 + 8 - 6 = 7 	ext{ sq. units}$.

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

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