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Find a Relation Between x and y Such that the Point (x, y) is Equidistant from the Point (3, 6) and (-3, 4)

Answer: Let the distance be q.

Using the distance formula,

q = √ (x2 – x1)2 + (y2 – y1)2

Let A = (x,y)

Similarly let us assume B = (3,6) and C = (-3,4)

So, distance between A and B = √ (x + 3)2 + (y – 4)2

Also, distance between A and C = √ (x – 3)2 + (y – 6)2

AB = AC

∴ √ (x + 3)2 + (y – 4)2 = √ (x – 3)2 + (y – 6)2

We square both sides

(x + 3)2 + (y – 4)2 = (x – 3)2 + (y – 6)2

⇒ x2 + 6x + 9 + y2 – 8y +16 = x2 – 6x + 9 + y2 – 12y + 36

⇒ – 8y + 6x + 25 = 6x – 12y + 45 

⇒ y + 3x = 5 

Therefore the relation established between the x and y coordinates is y + 3x = 5.