# 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.

#### 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