If the distance of p(x,y) from A(5,1) and B(-15) are equal then prove that 3x=2y.
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Let P (x, y) be equidistant from A ( 5, 1) and B ( – 1, 5).
Given, PA = PB
∴ PA2 = PB2
⇒ (x – 5)2 + (y –1)2 = {x – (– 1)}2 + (y – 5)2
⇒ (x – 5)2 + (y –1)2 = (x + 1)2 + (y – 5)2
⇒ x 2 + 25 –10x + y 2 + 1 – 2y = x 2 + 1 +2x + y 2 + 25 – 10y
⇒ –10x – 2y = 2x – 10y
⇒ 10y – 2y = 2x + 10x
⇒ 8y = 12x
⇒ 2y = 3x
Thus, 3 x = 2 y
Given, PA = PB
∴ PA2 = PB2
⇒ (x – 5)2 + (y –1)2 = {x – (– 1)}2 + (y – 5)2
⇒ (x – 5)2 + (y –1)2 = (x + 1)2 + (y – 5)2
⇒ x 2 + 25 –10x + y 2 + 1 – 2y = x 2 + 1 +2x + y 2 + 25 – 10y
⇒ –10x – 2y = 2x – 10y
⇒ 10y – 2y = 2x + 10x
⇒ 8y = 12x
⇒ 2y = 3x
Thus, 3 x = 2 y
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