Question

The distance between the parallel lines $y=x+a,y=x+b$ is:

• A. $\frac{\left|\left(b-a\right)\right|}{\sqrt{2}}$
• B. $\left|\left(a-b\right)\right|$
• C. $a+b$
• D. $\frac{\left|\left(b+a\right)\right|}{\sqrt{2}}$

Answer 1

The correct option is A

$\frac{\left|\left(b-a\right)\right|}{\sqrt{2}}$

Explanation for the correct option:

Finding the distance between the parallel lines:

The given equations are: $y=x+a,y=x+b$

Rewriting the given equations we get

$$\begin{gathered} x-y+a=0\dots \dots \left(1\right) \\ x-y+b=0\dots \dots \left(2\right) \end{gathered}$$

We know that the general equation of the line is $ax+by+c=0$

Since we have two given equations, we consider as $ax+by+c_1=0\dots \dots \left(3\right)$ and $ax+by+c_2=0\dots \dots \left(4\right)$

Comparing equations $\left(1\right)$ and $\left(3\right)$, $\left(2\right)$ and $\left(4\right)$ respectively we get

$a_1=1,b_1=-1,c_1=a,a_2=1,b_2=-1,c_2=b$

Distance between the two parallel lines will be:

$$\begin{gathered} \mathbf{=}\frac{\mathbf{}\left[(c_1-c_2)\right]}{\sqrt{\left({a_1}^2+{b_1}^2\right)}} \\ =\frac{\left|\left(a-b\right)\right|}{\sqrt{\left(1+1\right)}} \\ =\frac{\left|\left(a-b\right)\right|}{\sqrt{2}} \\ =\frac{\left|\left(b-a\right)\right|}{\sqrt{2}} \end{gathered}$$

Distance between the two parallel lines$=\frac{\left|\left(b-a\right)\right|}{\sqrt{2}}$

Hence, option (A) is the correct answer.