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The ratio in which the line 3x + 4y + 2 = 0 divides the distance between the lines 3x + 4y + 5 = 0 and 3x + 4y – 5 = 0 is ______. - Mathematics

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प्रश्न

The ratio in which the line 3x + 4y + 2 = 0 divides the distance between the lines 3x + 4y + 5 = 0 and 3x + 4y – 5 = 0 is ______.

विकल्प

  • 1 : 2

  • 3 : 7

  • 2 : 3

  • 2 : 5

MCQ
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उत्तर

The ratio in which the line 3x + 4y + 2 = 0 divides the distance between the lines 3x + 4y + 5 = 0 and 3x + 4y – 5 = 0 is 3 : 7.

Explanation:

The given equations are

3x + 4y + 5 = 0  .......(i)

3x + 4y – 5 = 0  .....(ii)

And 3x + 4y + 2 = 0  ......(iii)

Clearly, equation (i), (ii) and (iii) are parallel to each other as the coefficients of x and y are same.

Distance between parallel lines (i) and (iii) we get

`|(5 - 2)/sqrt((3)^2 + (4)^2)| = 3/5`  ......`[(because "Distance between two"),("parallel lines" = |("c"_1 - "c"_2)/sqrt("a"^2 + "b"^2)|)]`

Distance between parallel lines (ii) and (iii) we get

`|(-5 - 2)/sqrt((3)^2 + (4)^2)| = 7/5`

∴ Ratio between the distances = `3/5 : 7/5` = 3 : 7

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अध्याय 10: Straight Lines - Exercise [पृष्ठ १८३]

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एनसीईआरटी एक्झांप्लर Mathematics [English] Class 11
अध्याय 10 Straight Lines
Exercise | Q 40 | पृष्ठ १८३

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