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Question
Find equation of the line which is equidistant from parallel lines 9x + 6y – 7 = 0 and 3x + 2y + 6 = 0.
Solution
The equations of the given lines are
9x + 6y – 7 = 0 …(1)
3x + 2y + 6 = 0 …(2)
Let P (h, k) be the arbitrary point that is equidistant from lines (1) and (2). The perpendicular distance of P (h, k) from line (1) is given by
`d_1 = |9h + 6k - 7|/((9)^2 + (6)^2) = |9h + 6k - 7|/sqrt117 = |9h + 6k - 7|/(3sqrt13)`
The perpendicular distance of P (h, k) from line (2) is given by
`d_1 = |3h + 2k + 6|/((3)^2 + (2)^2) d_1 = |3h + 2k + 6|/sqrt13`
Since P (h, k) is equidistant from lines (1) and (2), d1 = d2
= `|9h + 6k - 7|/(3sqrt13) = |3h + 2k + 6|/sqrt13`
= |9h + 6k - 7| = 3|3h + 2k + 6|
= |9h + 6k - 7| = ±3|3h + 2k + 6|
= 9h + 6k - 7 = 3(3h + 2k + 6) or 9h + 6k - 7 = -3 (3h + 2k + 6)
The case 9h + 6k - 7 = 3 (3h + 2k + 6) is not possible as
9h + 6k - 7 3(3h + 2k + 6) = -7 = 18 (which is absurd)
∴ 9h + 6k - 7= -3(3h + 2k + 6)
9h + 6k - 7 = - 9h - 6k - 18
= 18h + 12k + 11 = 0
Thus, the required equation of the line is 18x + 12y + 11 = 0.
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