Question

find the coordinates of a point on x-are is which is equidistant from P (2,-5) and a(-2,9) P complete the following activity.​

Answer 1

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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).

Answer 2

Answer:

Step-by-step explanation:

Since point on x−axis, then coordinate of the point is (x,0).  

According to the question this point (x,0) is equidistant from the points (2,−5) and (−2,9).

That is, distance from  (x,0) and (2,−5) = distance from (x,0) and (−2,9).

(2−x)  2 +(−5−0)  2

 

​

=  

(−2−x)  2  +(9−0)  2

 

​

 

⇒(2−x)  2 +(−5)  2

=(−2−x)  2  +(9)  2

 

⇒4−4x+x  2

+25=4+4x+4+81

⇒8x=25−81

⇒8x=−56

⇒x=−7

So, point is (−7,0).