Using the method of integration,find the area | vedclass

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(A) The given equations of the lines are:$2x + y = 4$  ....... $(1)$$3x - 2y = 6$  ....... $(2)$$x - 3y + 5 = 0$  ....... $(3)$First,find the vertices

Vedclass · Accepted Answer

$\frac{7}{2}$

Vedclass · Answer

(A) The given equations of the lines are:$2x + y = 4$  ....... $(1)$$3x - 2y = 6$  ....... $(2)$$x - 3y + 5 = 0$  ....... $(3)$First,find the vertices of the triangle by solving the equations in pairs:Intersection of $(1)$ and $(2)$: $2x + y = 4$ and $3x - 2y = 6$. Solving gives $A = (2, 0)$ is incorrect based on the graph. Let's re-solve: $y = 4 - 2x$. Substituting into $(2)$: $3x - 2(4 - 2x) = 6 \implies 3x - 8 + 4x = 6 \implies 7x = 14 \implies x = 2, y = 0$. So $B = (2, 0)$.Intersection of $(1)$ and $(3)$: $2x + y = 4$ and $x - 3y = -5$. Solving gives $A = (1, 2)$.Intersection of $(2)$ and $(3)$: $3x - 2y = 6$ and $x - 3y = -5$. Solving gives $C = (4, 3)$.The area of the region bounded by the lines is the area of $\Delta ABC$. $AL$ and $CM$ are the perpendiculars on the $x$-axis.Area $(\Delta ABC) = 	ext{Area under } AC - 	ext{Area under } AB - 	ext{Area under } BC$Area $= \int_{1}^{4} \frac{x+5}{3} dx - \int_{1}^{2} (4-2x) dx - \int_{2}^{4} \frac{3x-6}{2} dx$$= \frac{1}{3} [\frac{x^2}{2} + 5x]_1^4 - [4x - x^2]_1^2 - \frac{1}{2} [\frac{3x^2}{2} - 6x]_2^4$$= \frac{1}{3} [(\frac{16}{2} + 20) - (\frac{1}{2} + 5)] - [(8 - 4) - (4 - 1)] - \frac{1}{2} [(\frac{3(16)}{2} - 24) - (\frac{3(4)}{2} - 12)]$$= \frac{1}{3} [28 - 5.5] - [4 - 3] - \frac{1}{2} [0 - (6 - 12)]$$= \frac{1}{3} [22.5] - 1 - \frac{1}{2} [6]$$= 7.5 - 1 - 3 = 3.5 = \frac{7}{2} 	ext{ sq. units}$.

Using the method of integration,find the area of the region bounded by the lines: $2x + y = 4$,$3x - 2y = 6$,and $x - 3y + 5 = 0$.

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