Question

Find the coordinates of the point equidistant from three given points A(5, 3), B(5, –5) and C(1, –5).

Answer 1

Let the required point be P (x, y). Then AP = BP = CP

That is, `(AP)^2 = (BP)^2 = (cp)^2`

This means`(Ap)^2 = (BP)^2`

`⇒(x-5)^2 +(y-3)^2 = (x-5)^2 +(y+5)^2`

`⇒x^2-10x+25+y^2-6y +9 =x^2-10x +25+y^2 +10y+25`

`⇒x^2 -10x +y^2 -6y +34 =x^2 - 10x+y^2+10y+50`

`⇒x^2-10x +y^2-6y-x^2 +10x-y^2-10y = 50-34`

⇒ -16y=16

`⇒y=-16/16=-1`

And `(BP)^2 = (CP)^2`

`⇒(x-5)^2 +(y+5)^2 = (x-1)^2 +(y+5)^2`

`⇒ x^2 -10x +25 +y^2 +10y +25 = x^2 -2x +1 +y^2 +10y +25`

`⇒ x^2 -10x +y^2 +10y + 50= x^2 -2x +y^2 +10y +26`

`⇒x^2 -10x  +y^2 +10y -x^2 +2x - y^2 -10y = 26-50`

⇒ -8x = -24 

`⇒ x = (-24)/(-8) = 3`

Hence, the required point is (3, -1 ).