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

**Solution:**

The distance between the two points can be measured using the Distance Formula which is given by: Distance Formula = √ [(x_{2 - }x_{1})^{2} + (y_{2} - y_{1})^{2}]

The point is 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 PA, 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

To find the distance between PB, 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

By the given condition, these distances are equal in measure. Hence, PA = PB

√(x - 2)² + (5)² = √(x + 2)² + (- 9)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 axis is (- 7, 0).

**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

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

Final Answer: The point equidistant from the given points (2, - 5) and (- 2, 9) on the axis is (- 7, 0)