The equation of the given lines are 9x +6y -7 =0........(1)

3x + 2y +6 = 0........(2)

Consider P(h, k) be the arbitrary point that is equidistant from lines (1) and (2) .

Here, the perpendicular distance of P(h, k) from line (1) is,

\( d_1=\dfrac{|9h+6k-7|}{(9)^2+(6)^2}\)

\(=\dfrac{|9h+6k-7|}{\sqrt{117}}\)

\( =\dfrac{|9h+6k-7|}{3\sqrt{13}}\)

= Similarly, the perpendicular distance of P(h, k) from line (2) is,

\( d_1=\dfrac{|3h+2k+6|}{\sqrt{(3)^2+(2)^2}}\)

\( =\dfrac{|3h+2k+6|}{\sqrt{13}}\)

We know that P(h, k) is equidistant from lines (1) and (2)

That is, d_{1}=d_{2}

Substitute the values,

\( \dfrac{|9h+6k-7|}{3\sqrt{13}}\)\( =\dfrac{|3h+2k+6|}{\sqrt{13}}\)

= |9h+6k-7|=3|3h+2k+6|

= |9h+6k-7|=\( \pm\)3|3h+2k+6|

Here

9h + 6k – 7 = 3 (3h + 2k + 6) or 9h + 6k – 7 = – 3 (3h + 2k + 6)

9h + 6k – 7 = 3 (3h + 2k + 6) is not possible as

9h + 6k – 7 = 3 (3h + 2k + 6)

By further calculation

– 7 = 18 (which is wrong)

We know that

9h + 6k – 7 = -3 (3h + 2k + 6)

By multiplication

9h + 6k – 7 = -9h – 6k – 18

We get

18h + 12k + 11 = 0

Hence, the required equation of the line is 18x + 12y + 11 = 0

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