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# 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 + 6 y - 7 = 0 ....(1)

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

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

Then the perpendicular distance of P (h, k) from the line (1) is given by

d₁ = |9h + 6k - 7|/√(9)² + (6)²

= |9h + 6k - 7|/√117

= |9h + 6k - 7|/3√13

And the perpendicular distance of P (h, k) from the line (2) is given by

d₂ = |3h + 2k + 6|/√(3)² + (2)²

= |3h + 2k + 6|/√13

Since P (h, k) is equidistant from lines (1) and (2), d₁ = d₂

Therefore,

|9h + 6k - 7|/3√13 = |3h + 2k + 6|/√13

⇒ |9h + 6k - 7| = 3|3h + 2k + 6|

⇒ 9h + 6k - 7 = ± 3(3h + 2k + 6)

__Case I__:

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

⇒ 9h + 6k - 7 = 9h + 6k + 18

⇒ 9h + 6k - 7 - 9h - 6k - 18 = 0

⇒ 25 = 0, which is absurd, hence this case is not possible.

__Case II__:

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 + 12 y + 11 = 0

NCERT Solutions Class 11 Maths Chapter 10 Exercise ME Question 21

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

**Summary:**

The equation of the line which is equidistant from parallel lines 9x + 6y - 7 = 0 and 3x + 2y + 6 = 0 is 18x + 12 y + 11 = 0

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