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

Let the point be P(x,y), then since all three points are equidistant therefore


PA = PB = PC


By distance formula, as shown below:



We have, PA = √{(5 – x)2 + (3 – y)2}


= √{25 + x2 – 10x + 9 + y2 – 6y}


⇒ PA = √{34 + x2 – 10x + y2 – 6y}


PB = √{(5 – x)2 + (– 5 – y)2}


= √{25 + x2 – 10x + 25 + y2 + 10y}


⇒ PB = √{50 + x2 – 10x + y2 + 10y}


PC = √{(1 – x)2 + (– 5 – y)2}


= √{1 + x2 – 2x + 25 + y2 + 10y}


⇒ PC = √{26 + x2 – 2x + y2 + 10y}


Squaring PA and PB we get


{34 + x2 – 10x + y2 – 6y} = {50 + x2 – 10x + y2 + 10y}


⇒ – 16 = 16y


⇒ y = – 1


Squaring PB and PC we get


{50 + x2 – 2x + y2 + 10y} = {26 + x2 – 10x + y2 + 10y}


24 = – 8x


x = – 3


P(– 3, – 1)


13