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

Asked by Abhisek | 1 year ago |  104

##### Solution :-

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)

PA = PB

Firstly let us calculate

Calculating PA

P ≡ (x, y, z) and A ≡ (1, 2, 3)

By using the formula,

Distance PA =

$$\sqrt{(x^2 – x^1)^2 + (y^2 – y^1)^2 + (z^2 – z^1)^2}$$

So here,

x1 = x, y1 = y, z1 = z

x2 = 1, y2 = 2, z2 = 3

Distance PA

$$\sqrt{(1 – x)^2 + (2 – y)^2 + (3 – z)^2}$$

Calculating PB

P ≡ (x, y, z) and B ≡ (3, 2, – 1)

By using the formula,

Distance PB =

$$\sqrt{(x^2 – x^1)^2 + (y^2 – y^1)^2 + (z^2 – z^1)^2}$$

So here,

x1 = x, y1 = y, z1 = z

x2 = 3, y2 = 2, z2 = – 1

Distance PB =

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

Since PA = PB

Square on both the 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

The required equation is x – 2z = 0

Answered by Pragya Singh | 1 year ago

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