# If A(–2, 2, 3) and B(13, –3, 13) Are Two Points. Find the Locus of a Point P Which Moves in Such a Way the 3pa = 2pb. - Mathematics

If A(–2, 2, 3) and B(13, –3, 13) are two points.
Find the locus of a point P which moves in such a way the 3PA = 2PB.

#### Solution

Let coordinates of point P be (xyz).
Given:
3PA = 2PB

$\Rightarrow 3\left( \sqrt{\left( x + 2 \right)^2 + \left( y - 2 \right)^2 + \left( z - 3 \right)^2} \right) = 2\left( \sqrt{\left( x - 13 \right)^2 + \left( y + 3 \right)^2 + \left( z - 13 \right)^2} \right)$
$\Rightarrow 3\left( \sqrt{x^2 + 4x + 4 + y^2 - 4y + 4 + z^2 - 6z + 9} \right) = 2\left( \sqrt{x^2 - 26x + 169 + y^2 + 6y + 9 + z^2 - 26z + 169} \right)$
$\text{ Squaring both sides },$
$\Rightarrow 9\left( x^2 + y^2 + z^2 + 4x - 4y - 6z + 17 \right) = 4\left( x^2 + y^2 + z^2 - 26x + 6y - 26z + 347 \right)$
$\Rightarrow 9 x^2 + 9 y^2 + 9 z^2 + 36x - 36y - 54z + 153 = 4 x^2 + 4 y^2 + 4 z^2 - 104x + 24y - 104z + 1388$
$\Rightarrow 5 x^2 + 5 y^2 + 5 z^2 + 140x - 60y + 50z - 1235 = 0$
$\Rightarrow 5\left( x^2 + y^2 + z^2 \right) + 140x - 60y + 50z - 1235 = 0$
$\therefore 5\left( x^2 + y^2 + z^2 \right) + 140x - 60y + 50z - 1235 = 0 \text{ is the locus of the point P } .$

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#### APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 28 Introduction to three dimensional coordinate geometry
Exercise 28.2 | Q 16 | Page 10