Question

find the point on the x axis which is the eqidistant from the points (-2,-5) and (-2,9)​

Answer 1

Solution :-

Let the point on the x - axis ( x, 0 ) . Which is the equidistant from the points ( -2,-5) and ( -2,9 )

Therefore,

According to the question,

The distance between ( x, 0) and ( 2 , 5) is equal to the distance between ( x,0) and ( -2 , 9 )

Therefore,

By using distance formula,

= √( x2 - x1 )^2 + ( y2 - y1 )^2

Let's compare the given points with

( x1 , y1 ) and ( x2 , y2 )

Put the required values,

√(2 - x)^2 + (5 - 0)^2 = √(-2-x)^2+(9-0)^2

[ By using identity ( a - b)^2 = a^2 + b^2

- 2ab and ( a + b)^2 = a^2 + b^2 + 2ab) ]

√ (4 + x^2 - 4x +25 ) = √(4+x^2 + 4x + 81 )

√( x^2 - 4x + 29 ) = √( x^2 + 4x + 85)

Now, squaring on both the sides,

( √x^2 - 4x + 29)^2 = ( √x^2 + 4x + 85)^2

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

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

-8x = 56

x = 56 / -8

x = -7

Hence, The point on x axis is

( -7 , 0 )$.$

Answer 2

Given,

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

To find,

The point on the x-axis which is equidistant from the two given points.

Solution,

Now,the point is on x-axis,so the value 'y' of the given point will be zero.

Let,the value of the 'x' value of the given point = x

[Assume,x as a variable to do the further mathematical calculations.]

So,the point is = (x,0)

As mentioned in the question, the two points are equidistant from (x,0).

So,

Distance between (x,0) and (2,-5)

= V(x-2)2+(+5)

Distance between () and (-2,9).

=/(x+2)+(0-92

Now,the points are equidistant.

So,

(x-2)*+(0+5)? = /(x+2)+(0-9)

(x-2)?+(0+5)? = (x+2)P+(0-9)?

x?-4x+4+25 = x+4x+4+81

x2-4x-x2-4x = 4+81-4-25

-8x = 56 =

x = -7

Hence, the point on x-axis is (-7,0).