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 Video Solution: 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:
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