Find the equation of the set of points which are equidistant from the points (1, 2, 3)

Question:

Find the equation of the set of points which are equidistant from the points (1, 2, 3) and (3, 2, –1).

Solution:

Let P (xy, z) be the point that is equidistant from points A(1, 2, 3) and B(3, 2, –1).

Accordingly, PA = PB

$\Rightarrow \mathrm{PA}^{2}=\mathrm{PB}^{2}$

$\Rightarrow(x-1)^{2}+(y-2)^{2}+(z-3)^{2}=(x-3)^{2}+(y-2)^{2}+(z+1)^{2}$

$\Rightarrow x^{2}-2 x+1+y^{2}-4 y+4+z^{2}-6 z+9=x^{2}-6 x+9+y^{2}-4 y+4+z^{2}+2 z+1$

$\Rightarrow-2 x-4 y-6 z+14=-6 x-4 y+2 z+14$

$\Rightarrow-2 x-6 z+6 x-2 z=0$

$\Rightarrow 4 x-8 z=0$

$\Rightarrow x-2 z=0$

Thus, the required equation is $x-2 z=0$.

Leave a comment

Close

Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now