Question

The line $L$ given by $\frac{x}{5}+\frac{y}{b}=1$ passes through the point $(13,32).$ The line $K$ is parallel to $L$ and has the equation $\frac{x}{c}+\frac{y}{3}=1.$ Then the equation of line which is equidistant from line $L$ and $K,$ is

• A. $4x-y-17=0$
• B. $8x-2y-17=0$
• C. $8x-2y+17=0$
• D. $4x-y+17=0$

Answer 1

The correct option is B $8x-2y-17=0$

$\frac{x}{5}+\frac{y}{b}=1$ passes through $(13,32).$
$\Rightarrow \frac{13}{5}+\frac{32}{b}=1$
$\Rightarrow b=-20$
Since, lines $K$ and $L$ are parallel,
$\therefore 4=-\frac{3}{c}$
$\Rightarrow c=\frac{-3}{4}$
$\therefore$ Lines are $L\equiv 4x-y-20=0$ and $K\equiv 4x-y+3=0$

Let the point on the line which is equidistant from $L$ and $K$ be $(x^{\prime },y^{\prime })$
$\left|\frac{4x^{\prime }-y^{\prime }-20}{\sqrt{16+1}}\right|=\left|\frac{4x^{\prime }-y^{\prime }+3}{\sqrt{16+1}}\right|$
Taking negative sign, we get
$\frac{4x^{\prime }-y^{\prime }-20}{\sqrt{16+1}}=-\left(\frac{4x^{\prime }-y^{\prime }+3}{\sqrt{16+1}}\right)$
or, $8x^{\prime }-2y^{\prime }-17=0$
Replacing $x^{\prime }$ by $x$ and $y^{\prime }$ by $y$,
Required line is $8x-2y-17=0$