Question

find the point on the x—axis which is equidistant from (2,-5)and(-2,9).​

Answer 1

Answer:

According to the question we have A(2,−5) and B(−2,9) Let the points be P(x,0). So, AP=PB and AP2=PB2 ⇒(x−2)2+(0+5)2 = (x+2)2+(0−9)2 ⇒x2+4−4x+25=x2+4+4x+81 ⇒x2+29−4x=x2+85+4x ⇒−4x−4x=85−29 ⇒−8x=56 ⇒x=−7 Hence, point on the x-axis which is equidistant from (2,−5) and (−2,9) is (−7,0).

Answer 2

Step-by-step explanation:

Given:-

The points are (2,-5)and(-2,9).

To find:-

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

Solution:-

Let the required point which is equidistant from the given points be P (x,0)

Given points are (2,-5)and(-2,9).

Let A(2,-5) and B(-2,9)

A_______P________B

AP = PB

Length of AP :-

Let (x1, y1)=(2,-5)=>x1=2 and y1=-5

Let (x2, y2)=P(x,0)=>x2=x and y2=0

We know that

The distance between two points (x1 ,y1 )and

(x2, y2) is √[(x2-x1)^2+(y2-y1)^2] units

=> AP =√[(x-2)^2+(0-5)^2]

=> AP = √(x-2)^2+(-5)^2]

=> AP =√(x^2-4x+4+25)

=> AP =√(x^2-4x+29) units --------(1)

Length of PB:-

Let (x1, y1)=P(x,0)=>x1=x and y1=0

Let (x2, y2)=B(-2,9)=>x2=-2 and y2=9

We know that

The distance between two points (x1 ,y1 )and

(x2, y2) is √[(x2-x1)^2+(y2-y1)^2] units

=> PB =√[(-2-x)^2+(9-0)^2]

=> PB= √(-x-2)^2+(9)^2]

=> PB =√(x^2+4x+4+81)

=> PB =√(x^2+4x+85) units --------(2)

(1)=(2)

√(x^2-4x+29)= √(x^2+4x+85)

On squaring both sides then

=> [√(x^2-4x+29)]^2= [√(x^2+4x+85)]^2

=>x^2-4x+29=x^2+4x+85

=> -4x+29=4x+85

=> -4x-4x =85-29

=>-8x=56

=>x=-56/8

=>x=-7

The value of x = -7

The point = (-7,0)

Answer:-

The point on the x-axis which is equidistant from the given points is (-7,0)

Used formulae:-

• The distance between two points (x1 ,y1 )and
• (x2, y2) is √[(x2-x1)^2+(y2-y1)^2] units
• (a+b)^2=a^2+2ab+b^2
• (a-b)^2=a^2-2ab+b^2