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

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

Answers (1)

best_answer

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 equilibrium with 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 let's 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, the locus of the point which is equidistant from A and B is x-2z=0.

Posted by

Pankaj Sanodiya

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