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Find the Equations of the Lines Through the Point of Intersection of the Lines X − Y + 1 = 0 and 2x − 3y + 5 = 0, Whose Distance from the Point(3, 2) is 7/5. - Mathematics

Answer in Brief

Find the equations of the lines through the point of intersection of the lines x − y + 1 = 0 and 2x − 3y+ 5 = 0, whose distance from the point(3, 2) is 7/5.

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Solution

The equations of the lines through the point of intersection of the lines x − y + 1 = 0 and 2x − 3y + 5 = 0 is given by
x − y + 1 + a(2x − 3y + 5) = 0
⇒ (1 + 2a)x  + y(−3a − 1) + 5a + 1 = 0                          .....(1)
The distance of the above line from the point is given by \[\frac{3\left( 2a + 1 \right) + 2\left( - 3a - 1 \right) + 5a + 1}{\sqrt{\left( 2a + 1 \right)^2 + \left( - 3a - 1 \right)^2}}\]

\[\therefore \frac{\left| 3\left( 2a + 1 \right) + 2\left( - 3a - 1 \right) + 5a + 1 \right|}{\sqrt{\left( 2a + 1 \right)^2 + \left( - 3a - 1 \right)^2}} = \frac{7}{5}\]

\[ \Rightarrow \frac{\left| 5a + 2 \right|}{\sqrt{13 a^2 + 10a + 2}} = \frac{7}{5}\]

\[ \Rightarrow 25 \left( 5a + 2 \right)^2 = 49\left( 13 a^2 + 10a + 2 \right)\]

\[ \Rightarrow 6 a^2 - 5a - 1 = 0\]

\[ \Rightarrow a = 1, - \frac{1}{6}\]

Substituting the value of a in (1),  we get
3x − 4y + 6 = 0 and 4x − 3y + 1 = 0

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APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 23 The straight lines
Exercise 23.19 | Q 11 | Page 131
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