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

x² + 4 - 4x + 25 = x² + 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

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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).

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