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

Answers (1)
P Pankaj Sanodiya

Given, two points A=(1, 2, 3) and B=(3, 2, –1).

Let the point P= (x,y,z) be a point which is equidistance from the points A and B.

so,

The distance PA= The distance PB

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

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

\left [ (x-1)^2-(x-3)^2 \right ]+\left [ (y-2)^2-(y-2)^2 \right ]+\left [ (z-3)^2-(z+1)^2 \right ]=0

Now lets apply the simplification property,

a^2-b^2=(a+b)(a-b)

\left [ (2)(2x-4) \right ]+0+\left [ (-4)(2z-2) \right ]=0

4x-8-8z+8=0

4x-8z=0

x-2z=0

Hence locus of the point which is equidistant from A and B is x-2z=0.

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