# Find a relation between x and y such that the point (x, y) is equidistant from the point (3, 6) and (- 3, 4)

**Solution:**

The distance between any two points can be measured using the distance formula which is given by √ [(x₂_{ - }x₁)^{2} + (y₂ - y₁)^{2}]

Let point P (x, y) be equidistant from points A (3, 6) and B (- 3, 4).

Since they are equidistant, PA = PB

Hence by applying the distance formula for PA = PB, we get

√(x - 3)² + (y - 6)² = √(x - (- 3))² + (y - 4)²

√(x - 3)² + (y - 6)² = √(x + 3)² + (y - 4)²

By squaring, we get

PA^{2} = PB^{2}

(x - 3)^{2} + (y - 6)^{2} = (x + 3)^{2} + (y - 4)^{2}

x^{2} + 9 - 6x + y^{2} + 36 - 12y = x^{2} + 9 + 6x + y^{2} + 16 - 8y

6x + 6x + 12y - 8y = 36 - 16 [On further simplifying]

12x + 4y = 20

3x + y = 5

3x + y - 5 = 0

Thus, the relation between x and y is given by 3x + y - 5 = 0

**☛ Check: **NCERT Solutions for Class 10 Maths Chapter 7

**Video Solution:**

## Find a relation between x and y such that the point (x, y) is equidistant from the point (3, 6) and (- 3, 4)

NCERT Class 10 Maths Solutions Chapter 7 Exercise 7.1 Question 10

**Summary:**

The relation between x and y such that the point (x, y) is equidistant from the point (3, 6) and (- 3, 4) is 3x + y - 5 = 0.

**☛ Related Questions:**

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