Q. 265.0( 2 Votes )

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

Answer :

Let A (1, 2, 3) & B (3, 2, – 1)


Let point P be (x, y, z)


Since it is given that point P(x, y, z) is equal distance from point A(1, 2, 3) & B(3, 2, – 1)


i.e. PA = PB


Calculating PA


P ≡ (x, y, z) and A ≡ (1, 2, 3)


Distance PA


Here,


x1 = x, y1 = y, z1 = z


x2 = 1, y2 = 2, z2 = 3


Distance PA


Calculating PB


P ≡ (x, y, z) and B ≡ (3, 2, – 1)


Distance PB


Here,


x1 = x, y1 = y, z1 = z


x2 = 3, y2 = 2, z2 = – 1



Since PA = PB


Squaring both sides, we get –


PA2 = PB2


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


(1 + x2 – 2x) + (4 + y2 – 4y) + (9 + z2 – 6z)


= (9 + x2 – 6x) + (4 + y2 - 4y) + (1 + z2 + 2z)


– 2x – 4y – 6z + 14 = – 6x - 4y + 2z + 14


4x – 8z = 0


x – 2z = 0


This is the required equation.

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