Question

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

Answer 1

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.