Question

A line $L_1$ passes through the point $(1,1)$ and $(2,0)$ and another line $L_2$ passes through $(\frac{1}{2},0)$ and perpendicular to $L_1.$ Then the area (in sq. units) of the triangle formed by the lines $L_1,$ $L_2$ 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 $\frac{25}{16}$

Slope of $L_1$ is $\frac{1-0}{1-2}=-1$
Equation of $L_1$ is
$y-1=-(x-1)$
$\Rightarrow x+y=2\cdots \left(1\right)$

Equation of $L_2$ is
$y-0=1(x-\frac{1}{2})$
$\Rightarrow 2x-2y=1\cdots \left(2\right)$

Equation of $y-$axis is $x=0\cdots \left(3\right)$

Solving $\left(1\right),\left(2\right)$ and $\left(3\right)$, we get the intersection points and hence, vertices as $(0,2)$, $(0,\frac{-1}{2})$ and $(\frac{5}{4},\frac{3}{4})$
Area $=\frac{1}{2}[0(\frac{-1}{2}-\frac{3}{4})+0(\frac{3}{4}-2)+\frac{5}{4}(2-\frac{-1}{2})]$
$=\frac{25}{16}$