Question

Find the point which is equidistant from the points (3,2) and (7,2)​

Answer 1

Answer:

(5,y) => y is a real number

Step-by-step explanation:

Assume the point equidistant from A(3,2) and B(7,2) to be C(x,y)

As C is equidistant from A and B, CA = CB

CA = $\sqrt{(x-3)^2+(y-2)^2}$  

CB = $\sqrt{(x-7)^2+(y-2)^2}$

Hence, $(x-3)^2 + (y-2)^2 = (x-7)^2 + (y-2)^2$

=> $(x-3)^2 = (x-7)^2$

=> x - 3 = ±( x - 7 )

It can't be "+" as it would become undefined, hence

x - 3 = 7 - x

=> 2x = 10

=> x = 5

Hence, the point equidistant from (3,2) and (7,2) is (5,y) where "y" is any real number.

Answer 2

The point equidistant from  (3,2) and (7,2)​ is (5,2)

Please check attachment...

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