# What is the distance between the lines 2x + 4y + 9 = 0 and x + 2y + 7 = 0?

*print*Print*list*Cite

### 2 Answers

The distance between two parallel lines expressed in the form ax + by + c1 = 0 and ax + by + c2 = 0 is given by |c2 - c1|/sqrt(a^2 + b^2)

Here we have the lines 2x + 4y + 9 = 0 and x + 2y + 7= 0

Multiply the equation of the second line by 2 to get 2(x + 2y + 7)= 0

=> 2x + 4y + 14 = 0

The two lines are now in the form given earlier:

2x + 4y + 9 = 0

2x + 4y + 14 = 0

The distance between the lines is |14 - 9|/sqrt(2^2 + 4^2)

=> 5/sqrt 20

=> 5/2*sqrt 5

=> (sqrt 5)/2

**The distance between the lines 2x + 4y + 9 = 0 and x + 2y + 7= 0 is (sqrt 5)/2**

These are parallel lines and the distance between them is 1 1/4.

Since y-intercept of first equation is -9/4 and y-intercept of second equation is -7/2 the differnce between these intercepts is 1 1/4.