# Find the distance between parallel lines

(i) 15x + 8y – 34 = 0 and 15x + 8y + 31 = 0

(ii) l (x + y) + p = 0 and l (x + y) – r = 0

**Solution:**

It is known that the distance (d) between parallel lines Ax + By + C\(_1\) = 0 and Ax + By + C\(_2\) = 0 is given by

d = |C\(_1\) - C\(_2\)|/√A² + B²

**(i)** The given parallel lines are 15x + 8y – 34 = 0 and 15x + 8y + 31 = 0

Here, A = 15, B = 8, C\(_1\) = -34 and C\(_2\) = 31

Therefore, the distance between the parallel lines is

d = |-34 - 31|/√(15)² + (8)²

= |- 65|/√289 units

= 65/17 units

**(ii)** The given parallel lines are l (x + y) + p = 0 and l (x + y) – r = 0

Here, A = B = l, C\(_1\) = p and C\(_2\) = - r

Therefore, the distance between the parallel lines is

d = |C\(_1\) - C\(_2\)|/√A² + B²

= |p + r|/√l² + l² units

= |p + r|/√2l² units

= |p + r|/ |l| √2 units

= (1/√2) |(p + r)/l| units

NCERT Solutions Class 11 Maths Chapter 10 Exercise 10.3 Question 6

## Find the distance between parallel lines (i) 15x + 8y – 34 = 0 and 15x + 8y + 31 = 0 (ii) l (x + y) + p = 0 and l (x + y) – r = 0

**Summary:**

The distance between given two parallel lines is,

i) 65/17 units ii) (1/√2) |(p + r)/l| units

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