Question

Find a point on x axis which is equidistant from( 6,3) and (3,0)

Answer 1

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Let the two points be A ( 6 , 3 ) and B ( 3 , 0 )

and the point be C ( x' , 0 )

AC = BC

( x' - 6 )² + ( - 3 ) ² = ( x' - 3 ) ² + ( 0 )

x'² - 12x' + 36 + 9 = x'² - 6x’ + 9

6x' = 36

x' = 6


So the point is C ( 6 , 0 )

Answer 2

Answer:

The point C=(6,0)

Step-by-step explanation:

Given : A point on x axis which is equidistant from( 6,3) and (3,0).

To find : The point?

Solution :

Let the two points be A( 6 , 3 ) and B( 3 , 0 )

and the point be C (x,0)

We have given that,

AC = BC as they are equidistant

$(x-6)^2+(0-3)^2=(x-3)^2+(0)^2$

$x^2-12x+36+9=x^2-6x+9$

$6x=36$

$x=6$

Therefore, The point C=(6,0)