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Question
Find the co-ordinates of the point equidistant from three given points A(5,3), B(5, -5) and C(1,- 5).
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Solution
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 ).
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