A line L passes through the points $(1,1)$and $(2,0)$ and another line $L^{\prime}$ passes through $\left(\frac{1}{2}, 0\right)$ and perpendicular to L.Then the area of the triangle formed by

the lines $L, L^{\prime}$ and $y-$ axis, is

$\textbf{Hint: Product of slopes of two perpendicular lines is -1.}$

$\textbf{Step 1: Find the equation of two lines.}$

$\text{Using two points form, equation of line L is given by,}$

$\Rightarrow y-0=\dfrac{0-1}{2-1}(x-2)$

$\Rightarrow y=-(x-2)$

$\Rightarrow x+y=2.......(1)$

$\text{Slope of L = - 1}$

$\text{L and }$ $\text{L' are perpendicular to each other}$

$\text{Slope of L' =}$ $\dfrac{-1}{-1}=1$

$\text{Using point slope form, equation of line L' is given by,}$

$\Rightarrow y-0=1(x-\dfrac{1}{2})$

$\Rightarrow 2x-2y=1 ......(2)$

$\textbf{Step 2: Find the required area.}$

$\text{Let, L and L' intersect each other at B.}$

$\text{Solving eq(1) and eq(2) , we get,}$

$\Rightarrow A(x,y)=\left(\dfrac{5}{4},\dfrac{3}{4}\right)$

$\text{The intersection points of lines L and L' with y-axis are B(0,2) and C}$$\left(0,-\dfrac{1}{2}\right)$ $\text{respectively.}$

$\text{Using distance formula,}$

$AB=\sqrt{\left(0-\dfrac{5}{4}\right)^2+\left(2-\dfrac{3}{4}\right)^2}=\dfrac{5}{4}\sqrt2$

$\text{And,}$

$AC=\sqrt{\left(\dfrac{5}{4}-0\right)^2+\left(\dfrac{3}{4}+\dfrac{1}{2}\right)^2}=\dfrac{5}{4}\sqrt2$

$\text{Area of the triangle}$ $=\dfrac{1}{2}\times AB\times AC=\dfrac{1}{2}\times\dfrac{5}{4}\sqrt2\times\dfrac{5}{4}\sqrt2\ sq.unit$

$=\dfrac{25}{16}\ sq.unit$