Question

A line L passes through the points (1, 1) and (2, 0) and another line L’ passes through $[\frac{1}{2},0]$ and perpendicular to L. Then the area of the triangle formed by the lines L, L’ and y –axis, is

• A. $\frac{15}{8}$
• B. $\frac{25}{4}$
• C. $\frac{25}{8}$
• D. $\frac{25}{16}$

Answer 1

The correct option is D $\frac{25}{16}$

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