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Home » Find the equation of the curve formed by the set of all points which are equidistant from the points A(-1, 2, 3) and B(3, 2, 1).
Find the equation of the curve formed by the set of all points which are equidistant from the points A(-1, 2, 3) and B(3, 2, 1).
CBSE | Class 11 | Exercise 26B | Maths | RS Aggarwal | Three Dimensional Geometry
Find the equation of the curve formed by the set of all points which are equidistant from the points A(-1, 2, 3) and B(3, 2, 1).

Answer:

Let us take,

C(x,y,z) point equidistant from points A(-1, 2, 3) and B(3, 2, 1).

∴ AC = BC

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

Squaring on both sides,

(x + 1)2+ (y – 2)2 + (z – 3)2 = (x – 3)2 + (y – 2)2 + (z – 1)2

x2 + 2x + 1 + y2 – 4y + 4 + z2 – 6z + 9 = x2 – 6x + 9 + y2 – 4y + 4 + z2 – 2z + 1 8x – 4z = 0

Equation of curve is 8x – 4z = 0

i

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