Math, asked by choudhryvaibhav3892, 11 months ago

Find the point on x-axis which is equidistant from the points (-2, 5) and (2,-3).

Answers

Answered by nidhi14300
3

Answer:

point on x axis is (-2,0)

Step-by-step explanation:

let the point on x-axis be x(x,0)

GIVEN:A(-2,5) and B(2 ,-3) are equidistant from p,so AX =BX then,

∴AX =BX = AX²=BX²

∴(AX)²=  (X+2)² + 25

∴(BX)²=  (X-2)²  = 9

∴X² + 4x + 4 + 25 = X² - 4X + 4 +9

          ∴8X = - 16

          ∴X=-2

HENCE, POINT ON X AXIS IS -2.0

Answered by nikitasingh79
4

Given : points (-2, 5) and (2,-3)

 

To prove : a point on the x-axis which is equidistant

 

Solution :  

The coordinates of every point on the x- axis are of the form (x,0)

Given : A( - 2 ,5) and B( 2 , - 3 ) are equidistant from P( x,0), AP  = BP .

For distance AP : x1 = - 2 ,y1 = 5 , x2 = x & ​y2 = 0

Distance of AP =√ (x- (-2)² + (0 - 5)²

[Distance Formula=√(x2 - x1)² + (y2 - y1)²]

AP = √( x  + 2 )² + (- 5 )²

= √x² + 2² + 2 × 2 × x + 5²

[(a + b)² = a² + b² + 2ab]

AP = √x² +4 +4x +25

AP = √x² + 4 x  + 29…………………(1)                

For distance BP : x1 = 2 , x2 = x & ​ y1 = - 3 , y2 = 0

BP =√ (x −2) ² + ( 0 −(−3))²

[Distance Formula = √(x2 - x1)² + (y2 - y1)²]

BP = √(x - 2 )² + ( 3 )²

BP= √x² + 4 - 4 x  + 9

[(a- b)² = a² + b² - 2ab]

BP= √x² - 4 x  + 13…………………..(2)                  

AP = BP [equidistant from P]

√x² + 4 x  + 29 = √x² - 4 x  + 13

[From eq. 1 and 2 ]

On squaring both sides

x² + 4 x  + 29 = x² - 4 x  + 13

x² - x²  + 4 x + 4 x=  + 13 - 29

8 x = - 16

x = -16/8

x = - 2

Hence , the point on x - axis is (-2 , 0) .    

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