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Find 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

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प्रश्न

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

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उत्तर

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 + 4+ 2 = 0 and 3x + 4y + 5 = 0 is given by

\[\frac{\left| 2 - 5 \right|}{\sqrt{3^2 + 4^2}}\]
\[ = \left| \frac{3}{\sqrt{25}} \right| = \frac{3}{5}\]

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

\[\frac{\left| 2 + 5 \right|}{\sqrt{3^2 + 4^2}}\]
\[ = \left| \frac{7}{\sqrt{25}} \right| = \frac{7}{5}\]

Hence, the ratio is given by

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

 
 
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अध्याय 23: The straight lines - Exercise 23.16 [पृष्ठ ११४]

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आर.डी. शर्मा Mathematics [English] Class 11
अध्याय 23 The straight lines
Exercise 23.16 | Q 6 | पृष्ठ ११४

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