What is the equation of the line midway between the lines 3x – 4y + 12 = 0 and 3x – 4y = 6?

Concept:

Distance between two parallel lines ax + by + c1 = 0 and ax + by + c2 = 0 is \(\rm \frac {|c_1 - c_2|}{\sqrt {a^2 + b^2}} \)

Calculation:

We have, 3x – 4y + 12 = 0 and 3x – 4y = 6⇒ 3x – 4y - 6 = 0

∴ Distance between them = \(\rm \frac {|12 - (-6)|}{ \sqrt {3^2 + (-4)^2} } \)

= 18/5

Let, equation of line midway between the given lines 3x - 4y + c = 0

As, it is in the middle, distance from line 3x – 4y + 12 = 0 will be \(\frac {\frac{18}{5}}{2} = \frac{9}{5}\)

Now, 9/5 = \(\rm \frac {|12 - c|}{ \sqrt {3^2 + (-4)^2} } \)

⇒ 12 - c = 9

⇒ c = 3

Hence, option (4) is correct.