Question:
Find equation of the line which is equidistant from parallel lines 9x + 6y - 7 = 0 and 3x + 2y + 6 = 0
Answer:
Given equation are:
9x + 6y - 7 = 0 .........1
3x + 2y + 6 = 0.........2
Now equation of line which is equidistant form both the lines is
|(9x + 6y - 7)/√(92 + 62 )| = |(3x + 2y + 6)/√(32 + 22 )|
=> |(9x + 6y - 7)/√(81+36 )| = |(3x + 2y + 6)/√(9 + 4 )|
=> |(9x + 6y - 7)/√117| = |(3x + 2y + 6)/√13 |
=> |(9x + 6y - 7)/√117| = |(3x + 2y + 6)/√13 |
=> |(9x + 6y - 7)/√(13*9)| = |(3x + 2y + 6)/√13 |
=> |(9x + 6y - 7)/(3*√13)| = |(3x + 2y + 6)/√13 |
=> |(9x + 6y - 7)/3| = |3x + 2y + 6|
=> (9x + 6y - 7)/3 = 3x + 2y + 6 and (9x + 6y - 7)/3 = -(3x + 2y + 6)
=> 9x + 6y - 7 = 3*(3x + 2y + 6) and 9x + 6y - 7 = -3*(3x + 2y + 6)
=> 9x + 6y - 7 = 9x + 6y + 18 and 9x + 6y - 7 = -9x - 6y - 18
Now 9x + 6y - 7 = 9x + 6y + 18 which is not possible.
Again
9x + 6y - 7 = -9x - 6y - 18
=> 9x + 6y - 7 + 9x + 6y + 18 = 0
=> 18x + 12y + 11 = 0
So required equation is: 18x + 12y + 11 = 0
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