Question

find the relation between X and Y such that the point X Y is equal distance from the point 7, 1 and 3, 5​

Answer 1

Step-by-step explanation:

(x-7)^2+(y-1)^2=(x-3)^2+(y-5)^2

x^2-14x+49+y^2-2y+1=x^2-6x+9+y^2-10y+25

{applying identity (a+b)^2}

you get ,

-14x+49-2y+1=-6x+9-10y+25

-6x+14x-10y+2y=49+1+-9-25

-6x+14x-10y+2y=50-34

8x-8y=16

taking 8 as common , we get,

8(x-y)=16

x-y=16/8

x-y=2

Therefore, the relation is x-y=2

Hope this may help you!!