Question

The equation of a straight line $L$ is $x+y=2$, and $L_1$ is another straight line perpendicular to $L$ and passes through the point $(\frac{1}{2},0)$. The area of the triangle (in square units), formed by the y-axis and the lines $L,L_1$ is

• A. $\frac{25}{8}$
• B. $\frac{25}{16}$
• C. $\frac{25}{4}$
• D. $\frac{25}{12}$

Answer 1

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

Given $L:x+y-2=0$.Slope of this line is $-1$.
$\therefore$ The slope of the line $L_1$ which passes through $(\frac{1}{2},0)$ and perpendicular to the line $L$ is $1$ .
The equation of the line $L_1$ is $x-y-\frac{1}{2}=0$.
On solving both the equations, we can find the intersection point $A$ of $L$ and $L_1$ which is $(\frac{5}{4},\frac{3}{4})$.
Line $L$ intersects $y$-axis at point $B(0,2)$ and $L_1$ intersects $y$-axis at point $C(0,\frac{-1}{2})$.
Perpendicular distance from $A$ to the line segment $BC$ is $\frac{5}{4}$ and the $BC$ length is $\frac{5}{2}$.
$\therefore$ The area of the triangle formed by these three points is $\left(\frac{1}{2}\right)\left(\frac{5}{2}\right)\left(\frac{5}{4}\right)=\frac{25}{16}$.