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The Ratio in Which the Line 3x + 4y + 2 = 0 Divides the Distance Between the Line 3x + 4y + 5 = 0 and 3x + 4y − 5 = 0 is - Mathematics

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Question

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

Options

  • 1: 2

  • 3: 7

  • 2: 3

  •  2: 5

MCQ
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Solution

Here, in all equations the coefficient of x is same.
It means all the lines have same slope
So, all the lines are parallel.
Now, the distance between the line 3x + 4y + 2 = 0 and 3x + 4y + 5 = 0 is given by

\[\frac{\left| 2 - 5 \right|}{\sqrt{3^2 + 4^2}}\]

\[ = \frac{3}{\sqrt{25}} = \frac{3}{5}\]

Again, the distance between the line 3x + 4y + 2 = 0 and 3x + 4y − 5 = 0 is given by:

\[\frac{\left| 2 + 5 \right|}{\sqrt{3^2 + 4^2}}\]

\[ = \frac{7}{25} = \frac{7}{5}\]

Hence, the ratio is given by

\[\frac{3}{5} : \frac{7}{5}\]

\[ = 3 : 7\]

Hence, the correct answer is option (b).

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Chapter 23: The straight lines - Exercise 23.21 [Page 135]

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RD Sharma Mathematics [English] Class 11
Chapter 23 The straight lines
Exercise 23.21 | Q 33 | Page 135

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