The line 2x + 3y = 12 meets the x-axis at A a | vedclass

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(B) The line $AB$ is $2x + 3y = 12$. The slope of $AB$ is $m_1 = -\frac{2}{3}$.Since the line $ED$ is perpendicular to $AB$,its slope $m_2 = \frac{3}{

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$\frac{23}{3}$ sq. units

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(B) The line $AB$ is $2x + 3y = 12$. The slope of $AB$ is $m_1 = -\frac{2}{3}$.Since the line $ED$ is perpendicular to $AB$,its slope $m_2 = \frac{3}{2}$.The equation of line $ED$ passing through $(5, 5)$ is $y - 5 = \frac{3}{2}(x - 5)$,which simplifies to $3x - 2y = 5$.To find point $E$,solve the system of equations $2x + 3y = 12$ and $3x - 2y = 5$. Multiplying the first by $2$ and the second by $3$,we get $4x + 6y = 24$ and $9x - 6y = 15$. Adding them gives $13x = 39$,so $x = 3$. Substituting $x = 3$ into $2x + 3y = 12$ gives $6 + 3y = 12$,so $y = 2$. Thus,$E = (3, 2)$.Point $C$ is the intersection of $3x - 2y = 5$ with the $x$-axis $(y=0)$,so $3x = 5$,$x = \frac{5}{3}$. Thus,$C = (\frac{5}{3}, 0)$.Point $B$ is the intersection of $2x + 3y = 12$ with the $y$-axis $(x=0)$,so $3y = 12$,$y = 4$. Thus,$B = (0, 4)$.The area of quadrilateral $OCEB$ can be calculated as the area of $\Delta OBC$ minus the area of $\Delta OEC$ is not correct; rather,it is the area of $\Delta OBC$ plus the area of $\Delta BEC$ is not correct. The quadrilateral $OCEB$ is composed of $\Delta OBC$ and $\Delta BEC$ is incorrect. The area is best calculated as Area$(\Delta OBC)$ - Area$(\Delta OEC)$.Area$(\Delta OBC)$ = $\frac{1}{2} 	imes OB 	imes OC$ is wrong. The vertices are $O(0,0), C(\frac{5}{3}, 0), E(3, 2), B(0, 4)$.Using the shoelace formula: Area = $\frac{1}{2} |(0 	imes 0 + \frac{5}{3} 	imes 2 + 3 	imes 4 + 0 	imes 0) - (0 	imes \frac{5}{3} + 0 	imes 3 + 2 	imes 0 + 4 	imes 0)| = \frac{1}{2} |(\frac{10}{3} + 12) - 0| = \frac{1}{2} 	imes \frac{46}{3} = \frac{23}{3}$ sq. units.

The line $2x + 3y = 12$ meets the $x$-axis at $A$ and the $y$-axis at $B$. The line through $(5, 5)$ perpendicular to $AB$ meets the $x$-axis,$y$-axis,and the line $AB$ at $C, D, E$ respectively. If $O$ is the origin,then the area of the quadrilateral $OCEB$ is:

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