A line L passes through the point (1,  1) and (2, 0). Another line M which is perpendicular to L passes through the point [1/2, 0]. Then the area of the triangle formed by the lines L, M and y-axis is

Let P ( 1,  1),  Q ( 2,  0)

Eq of L = 0

$\begin{array}{l}\therefore misx-y-\frac{1}{2}=0\\ \Rightarrow 2x-2y-1=0--\left(2\right)\\ Solving\left(1\right)\&\left(2\right)\\ xy1\\ \begin{array}{cccc}1& -2& 1& 1\\ -2& -1& 2& -2\end{array}\\ \Rightarrow \frac{x}{-1-4}=\frac{y}{-4+1}=\frac{1}{-2-2}\Rightarrow \frac{x}{-5}=\frac{y}{-3}=\frac{1}{-4}\left(x,y\right)=\left(\frac{5}{4},\frac{3}{4}\right)\\ Equationofy-axisisx=0--\left(3\right)\\ P.Iof\left(1\right)\&\left(3\right)\left(0,2\right)\\ P.Iof\left(2\right)\&\left(3\right)\left(0,\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)\end{array}$

$\begin{array}{l}A\left(\frac{5}{4},\frac{3}{4}\right)B\left(0,2\right)C\left(0,\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)\\ Area=\frac{1}{2}\left|\begin{array}{cc}\frac{5}{4}-0& \frac{5}{4}-0\\ \frac{3}{4}-2& \frac{3}{4}+\frac{1}{2}\end{array}\right|\\ \Rightarrow \frac{1}{2}\left|\begin{array}{cc}\frac{5}{4}& \frac{5}{4}\\ \frac{-5}{4}& \frac{5}{4}\end{array}\right|\Rightarrow \frac{1}{2}\left|\frac{25}{16}+\frac{25}{16}\right|\Rightarrow \frac{25}{16}\end{array}$