# Find the point on the x-axis which is equidistant from (2, - 5) and (- 2, 9)

**Solution:**

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

Let's assume a point P on the x-axis which is of the form P(x, 0).

We have to find a point on the x-axis which is equidistant from A (2, - 5) and B (- 2, 9).

To find the distance between P and A, substitute the values of P (x, 0) and A (2, - 5) in the distance formula.

PA = √(x - 2)² + (0 - (- 5))²

= √(x - 2)² + (5)² --------- (1)

To find the distance between P and B, substitute the values of P (x, 0) and B (- 2, 9) in the distance formula.

PB = √(x - (- 2))² + (0 - 9)²

= √(x + 2)² + (- 9)² ---------- (2)

By the given condition, these distances are equal in measure.

Hence, PA = PB

√(x - 2)² + (5)² = √(x + 2)² + (- 9)² [From equation (1) and (2)]

Squaring on both sides, we get

(x - 2)^{2} + 25 = (x + 2)^{2} + 81

x^{2} + 4 - 4x + 25 = x^{2} + 4 + 4x + 81

8x = 25 - 81

8x = - 56

x = - 7

Therefore, the point equidistant from the given points on the x-axis is (- 7, 0).

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

**Video Solution:**

## Find the point on the x-axis which is equidistant from (2, - 5) and (- 2, 9)

NCERT Class 10 Maths Solutions Chapter 7 Exercise 7.1 Question 7

**Summary:**

The point on the x-axis which is equidistant from (2, - 5) and (- 2, 9) is (- 7, 0).

**☛ Related Questions:**

- Find the distance between the following pairs of points:(i) (2, 3), (4, 1)(ii) (- 5, 7), (- 1, 3)(iii) (a, b), (- a, - b)
- Find the distance between the points (0, 0) and (36, 15). Can you now find the distance between the two towns A and B discussed in Section 7.2?
- Determine if the points (1, 5), (2, 3) and (- 2, - 11) are collinear.
- Check whether (5, - 2), (6, 4) and (7, - 2) are the vertices of an isosceles triangle.

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