• # question_answer 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 perpendicular to L. Then the area of the triangle formed by the lines $L,L'$ and y- axis, is                                      [RPET 1991] A)            15/8                                           B)            25/4 C)            25/8                                           D)            25/16

Here $L\equiv x+y=2$and${L}'\equiv 2x-2y=1$.                    Equation of y-axis is $x=0$                    Hence the vertices of the triangle are $A(0,\,2),B\left( 0,-\frac{1}{2} \right)$ and$C\left( \frac{5}{4},\frac{3}{4} \right)$. Therefore, the area of the triangle is $\frac{1}{2}\left| \,\begin{matrix} 0 & 2 & 1 \\ 0 & -\frac{1}{2} & 1 \\ \frac{5}{4} & \frac{3}{4} & 1 \\ \end{matrix}\, \right|=\frac{25}{16}$.