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

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(C) The line $2x + 3y = 12$ meets the $y$-axis at $B$ where $x=0$,so $B = (0, 4)$.The slope of the line $2x + 3y = 12$ is $m_1 = -\frac{2}{3}$.The slo

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

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(C) The line $2x + 3y = 12$ meets the $y$-axis at $B$ where $x=0$,so $B = (0, 4)$.The slope of the line $2x + 3y = 12$ is $m_1 = -\frac{2}{3}$.The slope of the line perpendicular to $AB$ is $m_2 = -\frac{1}{m_1} = \frac{3}{2}$.The equation of the line passing through $(5, 5)$ with slope $\frac{3}{2}$ is $y - 5 = \frac{3}{2}(x - 5)$,which simplifies to $3x - 2y = 5$.This line meets the $x$-axis at $C$ where $y=0$,so $3x = 5 \implies C = (\frac{5}{3}, 0)$.To find $E$,we solve the system of equations:$2x + 3y = 12$ $(i)$$3x - 2y = 5$ (ii)Multiplying $(i)$ by $2$ and (ii) by $3$: $4x + 6y = 24$ and $9x - 6y = 15$.Adding these gives $13x = 39 \implies x = 3$. Substituting $x=3$ into $(i)$,$6 + 3y = 12 \implies 3y = 6 \implies y = 2$. So $E = (3, 2)$.The area of quadrilateral $OCEB$ can be calculated by splitting it into $\Delta OCE$ and $\Delta OEB$.Area of $\Delta OCE = \frac{1}{2} |x_O(y_C - y_E) + x_C(y_E - y_O) + x_E(y_O - y_C)| = \frac{1}{2} |0 + \frac{5}{3}(2 - 0) + 3(0 - 0)| = \frac{5}{3}$.Area of $\Delta OEB = \frac{1}{2} |x_O(y_E - y_B) + x_E(y_B - y_O) + x_B(y_O - y_E)| = \frac{1}{2} |0 + 3(4 - 0) + 0| = 6$.Total area $= \frac{5}{3} + 6 = \frac{23}{3} 	ext{ sq. units}$.

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

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