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(D) The equation of line $L$ passing through $(3,0)$ and $(1, 4/3)$ is given by: $y - 0 = \frac{4/3 - 0}{1 - 3}(x - 3)$ $y = -\frac{2}{3}(x - 3)$ $\Ri

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$39$

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(D) The equation of line $L$ passing through $(3,0)$ and $(1, 4/3)$ is given by: $y - 0 = \frac{4/3 - 0}{1 - 3}(x - 3)$ $y = -\frac{2}{3}(x - 3)$ $\Rightarrow 3y = -2x + 6$ $\Rightarrow 2x + 3y = 6$ (Eq. $i$) The slope of line $L$ is $m_1 = -2/3$. The slope of the line perpendicular to $L$ is $m_2 = -1/m_1 = 3/2$. The equation of the line passing through $(8,1)$ with slope $3/2$ is: $y - 1 = \frac{3}{2}(x - 8)$ $\Rightarrow 2y - 2 = 3x - 24$ $\Rightarrow 3x - 2y = 22$ (Eq. $ii$) Solving Eq. $i$ and Eq. $ii$ for the intersection point: $2x + 3y = 6$ (multiply by $2$) $\Rightarrow 4x + 6y = 12$ $3x - 2y = 22$ (multiply by $3$) $\Rightarrow 9x - 6y = 66$ Adding these gives $13x = 78 \Rightarrow x = 6$. Substituting $x=6$ into $2x + 3y = 6$ gives $12 + 3y = 6$ $\Rightarrow 3y = -6$ $\Rightarrow y = -2$. The intersection point is $(6, -2)$. The line $2x + 3y = 6$ intersects the $Y$-axis $(x=0)$ at $(0, 2)$. The line $3x - 2y = 22$ intersects the $Y$-axis $(x=0)$ at $(0, -11)$. The vertices of the triangle are $(6, -2)$,$(0, 2)$,and $(0, -11)$. The base of the triangle on the $Y$-axis is $|2 - (-11)| = 13$. The height of the triangle is the absolute value of the $x$-coordinate of the intersection point,which is $|6| = 6$. Area $= \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes 13 	imes 6 = 39 	ext{ sq units}$.

$A$ triangle is formed by the $Y$-axis,the straight line $L$ passing through the points $(3,0)$ and $(1, 4/3)$,and the straight line perpendicular to the line $L$ and passing through the point $(8,1)$. The area of the triangle (in sq units) is

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