Distance between the straight lines x + 2y = 5 and 2x + 4y = 11 is

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1. $$\dfrac{1}{\sqrt5}$$
2. $$\dfrac{1}{2\sqrt5}$$
3. $$\sqrt5$$
4. $$\dfrac{\sqrt5}{2}$$

Option 2 : $$\dfrac{1}{2\sqrt5}$$

Detailed Solution

Concept:

Condition for parallel lines: If the two lines are parallel, then their general forms of equations will differ only in the constant term and they will have the same coefficients of x and y.

That is,

ax + by + c1 = 0      ----(a)

ax + by + c2 = 0      ----(b)

Distance between two parallel lines: Distance d between two parallel lines ax + by + c1 = 0 and ax + by + c2 = 0 is given by,

$$d \ = \ \frac{\left | c_{1} \ - \ c_{2} \right |}{\sqrt{a^{2} \ + \ b^{2}}}$$      ----(A)

Calculation:

We have,

x + 2y = 5      ----(1)

Comparing it with the standard equation of a straight line i.e., ax + by + c = 0, we get

a1 = 1, b1 = 2, c1 = -5

And, 2x + 4y = 11

⇒ x + 2y = 11/2      ----(2)

Comparing it with the standard equation of a straight line i.e., ax + by + c = 0, we get

a2 = 1, b2 = 2, c2 = -11/2

Thus, the coefficients of x and y in equations (1) & (2) are the same.

That is, (a1 = a2) and (b1 = b2)

Thus, these are parallel lines.

Now, we have a = 1, b = 2, c1 = -5, and c2 = -11/2

$$\Rightarrow d \ = \ \frac{\left | -5 \ - \ (-11/2) \right |}{\sqrt{1^{2} \ + \ 2^{2}}}$$      [using (A)]

$$\Rightarrow d \ = \ \frac{\left | -5 \ + \ 11/2 \right |}{\sqrt{1 \ + \ 4}}$$

$$\Rightarrow d \ = \ \frac{1}{2\sqrt{5}}$$

Hence, distance between the straight lines x + 2y = 5 and 2x + 4y = 11 is $$\dfrac{1}{2\sqrt5}$$.
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