Question

Let $\alpha$ be the distance between the lines $-x+y=2$ 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 the above

Answer 1

The correct option is A

$20\sqrt{2}\beta =11\alpha$

Explanation for the correct option:

Finding the relation between $\alpha \&\beta$:

Step 1: The given equation of lines

$$\begin{gathered} \Rightarrow –x+y=2 \\ \Rightarrow y=x+2......\left(i\right) \end{gathered}$$

And

$$\begin{gathered} \Rightarrow x–y=2 \\ \Rightarrow y=x–2......\left(ii\right) \end{gathered}$$

Then we get the slope, $m=1$ for both the equation after comparing the general equation $y=mx+c$

Since slopes are the same, so the given lines are parallel.

Similarly, the lines

$$\begin{gathered} \Rightarrow 4x–3y=5 \\ \Rightarrow y=\frac{4}{3}x+\frac{5}{3}......\left(iii\right) \end{gathered}$$

And

$$\begin{gathered} \Rightarrow 6y–8x=1 \\ \Rightarrow y=\frac{8}{6}x+\frac{1}{6} \\ \Rightarrow y=\frac{4}{3}x+\frac{1}{6}......\left(iv\right) \end{gathered}$$

These are also parallel as the slopes of both equations are the same.

Step 2: Finding the distance between the two lines

We know that formula of the distance between two parallel lines,

$d=\frac{\left|\right(c_2–c_1\left)\right|}{\sqrt{(a^2+b^2)}}$

The distance between the lines $\left(i\right)\&\left(ii\right)$

$\begin{array}{rcl}\alpha & =& \frac{\left|\right(-2-2\left)\right|}{\sqrt{\left(1^2+1^2\right)}}\\ & =& \frac{\left|-4\right|}{\sqrt{\left(1+1\right)}}\\ & =& \frac{4}{\sqrt{2}}\\ \alpha & =& 2\sqrt{2}\end{array}$

Similarly the distance between the lines $\left(iii\right)\&\left(iv\right)$

$\begin{array}{rcl}\beta & =& \frac{\left|\left({\displaystyle \frac{-5}{3}}-{\displaystyle \frac{1}{6}}\right)\right|}{\sqrt{\left({\left({\displaystyle \frac{4}{3}}\right)}^2+1^2\right)}}\\ & =& \frac{\left|{\displaystyle \frac{-10-1}{6}}\right|}{\sqrt{\left({\displaystyle \frac{16}{9}}+1\right)}}\\ & =& \frac{\left|{\displaystyle \frac{-11}{6}}\right|}{\sqrt{\left({\displaystyle \frac{25}{9}}\right)}}\\ & =& \frac{{\displaystyle \frac{11}{6}}}{{\displaystyle \frac{5}{3}}}\\ & =& \frac{11}{6}\times \frac{3}{5}\\ \beta & =& \frac{11}{10}\end{array}$

Step 3: Finding the relation between $\alpha \&\beta$

$\begin{array}{rcl}\frac{\alpha }{\beta }& =& \frac{2\sqrt{2}}{{\displaystyle \frac{11}{10}}}\left[\because \alpha =2\sqrt{2}\text{and}\beta =\frac{11}{10}\right]\\ & =& \frac{20\sqrt{2}}{11}\\ 20\sqrt{2}\beta & =& 11\alpha \end{array}$

Hence, option (A) is the correct answer.