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

**Solution:**

Let A(1, 2, 3) and B (3, 2, - 1)

Let point P be (x, y, z)

Since it is given that point P (x, y, z) is equidistant from the points A(1, 2, 3) and B (3, 2, - 1) i.e., PA = PB

Firstly, let us calculate distances PA and PB

__Calculating PA__:

P (x, y, z) and A(1, 2, 3)

By using the distance formula,

Distance = √(x₂ - x₁)² + (y₂ - y₁)² + (z₂ - z₁)²

Using this,

PA = √(1 - x)² + (2 - y)² + (3 - z)²

__Calculating PB__:

P (x, y, z) and B (3, 2, - 1)

By using the distance formula,

Distance = √(x₂ - x₁)² + (y₂ - y₁)² + (z₂ - z₁)²

Using this,

PB = √(3 - x)² + (2 - y)² + (- 1 - z)²

Since, PA = PB

On squaring both the sides, we get

PA² = PB²

Therefore,

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

4x - 8z = 0

x - 2z = 0

Thus, the required equation is x - 2z = 0

NCERT Solutions Class 11 Maths Chapter 12 Exercise 12.2 Question 4

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

**Summary:**

The equation of the set of points which are equidistant from the points (1, 2, 3) and (3, 2, - 1) is x - 2z = 0

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