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Show that the Path of a Moving Point Such that Its Distances from Two Lines 3x − 2y = 5 and 3x + 2y = 5 Are Equal is a Straight Line. - Mathematics

Short Note

Show that the path of a moving point such that its distances from two lines 3x − 2y = 5 and 3x + 2y = 5 are equal is a straight line.

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Solution

Let P(hk) be the moving point such that it is equidistant from the lines 3x − 2y = 5 and 3x + 2y = 5

\[\left| \frac{3h - 2k - 5}{\sqrt{3^2 + 2^2}} \right| = \left| \frac{3h + 2k - 5}{\sqrt{3^2 + 2^2}} \right|\]
\[ \Rightarrow \left| 3h - 2k - 5 \right| = \left| 3h + 2k - 5 \right|\]
\[ \Rightarrow 3h - 2k - 5 = \pm \left( 3h + 2k - 5 \right)\]
\[ \Rightarrow 3h - 2k - 5 = 3h + 2k - 5 and 3h - 2k - 5 = - \left( 3h + 2k - 5 \right)\]
\[ \Rightarrow k = 0 \text{ and }  3h = 5\]

Hence, the path of the moving points are \[3x = 5 \text{ or }  y = 0\]  These are straight lines.

 
 
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APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 23 The straight lines
Exercise 23.15 | Q 13 | Page 108
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