Question

A line $L$ is drawn from $P(4,3)$ to meet the lines $L_1$ and $L_2$ given by $3x+4y+5=0$ and $3x+4y+15=0$ at points $A$ and $B$ respectively. From $A$, a line perpendicular to $L$ is drawn meeting the line $L_2$ at $A_1$. Similarly, from point $B$, a line perpendicular to $L$ is drawn meeting the line $L_1$ at $B_1$. Thus a parallelogram $A{A}_{1}B{B}_1$ is formed. Then the equation of $L$ so that the area of the parallelogram $A{A}_{1}B{B}_1$ is least is

• A. $x-7y+17=0$
• B. $7x+y+31=0$
• C. $x-7y-17=0$
• D. $x+7y-31=0$

Answer 1

The correct option is A $x-7y+17=0$

The given lines $(L_1$ and $L_2)$ are parallel and distance between them (BC or AD) is $(15-5)/5=2$ units.

Let $\angle BCA=\theta$

$\Rightarrow AB=BC\text{cosec}\theta =2\text{cosec}\theta$ and $A{A}_1=AD\sec \theta =2\sec \theta .$

Now area of parallelogram $A{A}_{1}B{B}_1$ is
$\Delta =AB\times A{A}_1=4\sec \theta \text{cosec}\theta$
$=\frac{8}{\sin 2\theta }$
Clearly, $\Delta$ is least for $\theta =\frac{\pi }{4}$.

Let slope of $AB$ be $m$.
Then, $1=\left|\frac{m+\frac{3}{4}}{1-\frac{3m}{4}}\right|$
$\Rightarrow 4m+3=\pm (4-3m)\Rightarrow m=\frac{1}{7}$ or $-7$
Hence, the equation of 'L' is
$x-7y+17=0$ or $7x+y-31=0$