Question

A line $L_1:2x-2y+5=0$ is rotated about its point of intersection with $y-$axis such that $L_1$ becomes $L_2$ and area of the triangle formed by $L_1,L_2$ and $x=4$ is $13\text{sq. unit}$, then $L_2$ will be

• A. $21x-8y+20=0$
• B. $5x+8y=20$
• C. $5x+8y+20=0$
• D. $21x-8y=20$

Answer 1

Answer: (A), (B)

The correct options are
A $21x-8y+20=0$
B $5x+8y=20$

$y-$ intercept of the $L_1$
$2\times 0-2y+5=0$
$y=\frac{5}{2}$ So,
$A=(0,\frac{5}{2})$

Intersection point of the $L_1$ with the line $x=4$
$2\times 4-2y+5=0y=\frac{13}{2}$
So, $B=(4,\frac{13}{2})$


Let the point $C$ on the line $L_2$ and $x=4$ be,
$C=(4,k)$
So the area of the triangle will be,
$A=\frac{1}{2}\left|\left|\begin{array}{ccc}1& 0& \frac{5}{2}\\ 1& 4& \frac{13}{2}\\ 1& 4& k\end{array}\right|\right|$
$k=0,13$

For $k=0$ the line will be,
$\frac{x}{4}+\frac{2y}{5}=1\Rightarrow 5x+8y=20$

For $k=13$ the line will be,
$\frac{y-13}{x-4}=\frac{13-\frac{5}{2}}{4}\Rightarrow 21x-8y+20=0$