Question

Let $a$ be the distance between line $ -x+y=2 $ and $ x-y=2 $ and $ \beta $ be the distance between the lines $ 4x-3y=5 $ and $ 6y-8x=1, $ then

• A. $ 20\sqrt{2}\beta =11\alpha $
• B. $ 20\sqrt{2}\alpha =11\beta $
• C. $ 11\sqrt{2}\beta =20\alpha $
• D. None of these

Answer 1

Answer: (A)

Given $\alpha$ be the distance between lines
$x-y+2=0$ and $x-y-2=0$.
$\therefore \alpha=\frac{|2+2|}{\sqrt{1+1}}=\frac{|4|}{\sqrt{2}}=2 \sqrt{2}$
and $\beta$ be the distance between the lines
$4 x-3 y-5=0$ and $4 x-3 y+1 / 2=0$
$\therefore \beta=\frac{\left|5+\frac{1}{2}\right|}{\sqrt{(4)^{2}+(3)^{2}}}$
$=\frac{|11|}{2 \sqrt{25}}=\frac{11}{10}$
Now, $\frac{\alpha}{\beta}=\frac{2 \sqrt{2}}{11 / 10}$
$\Rightarrow \frac{\alpha}{\beta}=\frac{20 \sqrt{2}}{11}$
$\Rightarrow 20 \sqrt{2} \beta=11 \alpha$