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Given a family of lines a (2x + y + 4) + b(x – 2y – 3) = 0, the number of lines belonging to the family at a distance 10 from P(2, –3) is ______.

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Question

Given a family of lines a (2x + y + 4) + b(x – 2y – 3) = 0, the number of lines belonging to the family at a distance `sqrt(10)` from P(2, –3) is ______.

Options

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  • 1

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MCQ
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Solution

Given a family of lines a (2x + y + 4) + b(x – 2y – 3) = 0, the number of lines belonging to the family at a distance `sqrt(10)` from P(2, –3) is 1.

Explanation:

The length of perpendicular from P(2, –3) on the given family of lines

= `(a(4 - 3 + 4) + b(2 + 6 - 3))/(sqrt((2a + b)^2 + (a - 2b)^2)) = ±sqrt(10)` ...(given)

`\implies` 5a + 5b = `±sqrt(10(5a^2 + 5b^2)`

`\implies` 25(a + b)2 = 50(a2 + b2)

`\implies` 25(a – b)2 = 0

`\implies` a = b

For which we get only line 3x – y + 1 = 0

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