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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