A line L passes through the points (1, 1) and | vedclass

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(D) The slope of line $L$ passing through $(1, 1)$ and $(2, 0)$ is $m = \frac{0 - 1}{2 - 1} = -1$.Equation of line $L$ is $y - 0 = -1(x - 2) \implies

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$\frac{25}{16}$

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(D) The slope of line $L$ passing through $(1, 1)$ and $(2, 0)$ is $m = \frac{0 - 1}{2 - 1} = -1$.Equation of line $L$ is $y - 0 = -1(x - 2) \implies x + y = 2$.Since $L'$ is perpendicular to $L$,its slope is $m' = -\frac{1}{m} = 1$.Equation of line $L'$ passing through $\left( \frac{1}{2}, 0 ight)$ is $y - 0 = 1\left( x - \frac{1}{2} ight) \implies x - y = \frac{1}{2} \implies 2x - 2y = 1$.The $y$-axis is $x = 0$.Intersection of $L$ and $y$-axis: $0 + y = 2 \implies y = 2$. Vertex $A = (0, 2)$.Intersection of $L'$ and $y$-axis: $2(0) - 2y = 1 \implies y = -\frac{1}{2}$. Vertex $B = (0, -\frac{1}{2})$.Intersection of $L$ and $L'$: Adding $x + y = 2$ and $x - y = \frac{1}{2}$ gives $2x = \frac{5}{2} \implies x = \frac{5}{4}$. Then $y = 2 - \frac{5}{4} = \frac{3}{4}$. Vertex $C = \left( \frac{5}{4}, \frac{3}{4} ight)$.The area of the triangle with vertices $(0, 2), (0, -\frac{1}{2}), \left( \frac{5}{4}, \frac{3}{4} ight)$ is $\frac{1}{2} 	imes 	ext{base} 	imes 	ext{height}$.Base on $y$-axis = $|2 - (-\frac{1}{2})| = \frac{5}{2}$.Height = $x$-coordinate of $C = \frac{5}{4}$.Area = $\frac{1}{2} 	imes \frac{5}{2} 	imes \frac{5}{4} = \frac{25}{16}$.

$A$ line $L$ passes through the points $(1, 1)$ and $(2, 0)$ and another line $L'$ passes through $\left( \frac{1}{2}, 0 \right)$ and is perpendicular to $L$. Then the area of the triangle formed by the lines $L, L'$ and the $y$-axis is:

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