Let L_1 be a line passing through (2,1) and ( | vedclass

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(A) Line $L_1$ passes through $(2,1)$ and $(3, \frac{5}{2})$.Slope of $L_1$ is $m_1 = \frac{\frac{5}{2} - 1}{3 - 2} = \frac{3/2}{1} = \frac{3}{2}$.The

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$\frac{121}{39}$

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(A) Line $L_1$ passes through $(2,1)$ and $(3, \frac{5}{2})$.Slope of $L_1$ is $m_1 = \frac{\frac{5}{2} - 1}{3 - 2} = \frac{3/2}{1} = \frac{3}{2}$.The equation of $L_1$ is $(y - 1) = \frac{3}{2}(x - 2)$,which simplifies to $3x - 2y = 4$.Line $L_2$ is perpendicular to $L_1$,so its slope $m_2 = -\frac{1}{m_1} = -\frac{2}{3}$.$L_2$ passes through $(4, -1)$,so its equation is $(y + 1) = -\frac{2}{3}(x - 4)$,which simplifies to $2x + 3y = 5$.$L_1$ intersects the $y$-axis at $x=0$,so $-2y = 4 \Rightarrow y = -2$. Point is $(0, -2)$.$L_2$ intersects the $y$-axis at $x=0$,so $3y = 5 \Rightarrow y = \frac{5}{3}$. Point is $(0, \frac{5}{3})$.To find the intersection of $L_1$ and $L_2$,solve $3x - 2y = 4$ and $2x + 3y = 5$. Multiplying the first by $3$ and second by $2$: $9x - 6y = 12$ and $4x + 6y = 10$. Adding them gives $13x = 22 \Rightarrow x = \frac{22}{13}$. Substituting $x$ gives $y = \frac{7}{13}$.The vertices of the triangle are $(0, -2)$,$(0, \frac{5}{3})$,and $(\frac{22}{13}, \frac{7}{13})$.Area $= \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes |\frac{5}{3} - (-2)| 	imes |\frac{22}{13}| = \frac{1}{2} 	imes \frac{11}{3} 	imes \frac{22}{13} = \frac{121}{39}$.

Let $L_1$ be a line passing through $(2,1)$ and $(3, \frac{5}{2})$. $L_2$ is a line perpendicular to $L_1$ and passing through $(4,-1)$. The area of the triangle formed by $L_1$,$L_2$ and the $y$-axis is

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