Find the point on the x axis which is equidistant from(2,-5) and (-2,9)
Answers
Step-by-step explanation:
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