Question

Question 21 Find equation of the line which is equidistant from parallel lines 9x + 6y – 7 = 0 and 3x + 2y + 6 = 0.

Class X1 - Maths -Straight Lines Page 234

Answer 1

Let two line ax + by + c = 0
and, ax + by+ k = 0 are parallel .
then, distance between these line = |c-k|/√(a²+b²)


Here,
9x + 6y -7 = 0
3(3x + 2y - 7/3) = 0
3x + 2y -7/3 = 0-----(i)
and 3x + 2y + 6 = 0 .---(ii)

Let equation of line parallel to either equation (i) or equation (ii) is
3x + 2y + k = 0 -----(iii)

Now,
A/c to question,
Given line is equidistant from lines (i) and (ii)
So,

|k -(-7/3)|/√(3²+2²) = |k-6|/√(3²+2²)
|k+7/3| = |k-6|
k+7/3 =±(k-6)
k + 7/3 = -k+6
2k = 6-7/3 = 11/3
k = 11/6

Hence, equation of required line is 3x + 2y + 11/6 = 0
Or,
18x + 12y + 11 = 0

Answer 2

Answer:

Let two line ax + by + c = 0

and, ax + by+ k = 0 are parallel .

then, distance between these line = |c-k|/√(a²+b²)

Here,

9x + 6y -7 = 0

3(3x + 2y - 7/3) = 0

3x + 2y -7/3 = 0-----(i)

and 3x + 2y + 6 = 0 .---(ii)

Let equation of line parallel to either equation (i) or equation (ii) is

3x + 2y + k = 0 -----(iii)

Now,

A/c to question,

Given line is equidistant from lines (i) and (ii)

So,

|k -(-7/3)|/√(3²+2²) = |k-6|/√(3²+2²)

|k+7/3| = |k-6|

k+7/3 =±(k-6)

k + 7/3 = -k+6

2k = 6-7/3 = 11/3

k = 11/6

Hence, equation of required line is 3x + 2y + 11/6 = 0

Or,

18x + 12y + 11 =