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

Question:

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

Solution:

Let the required point be P(xy). Then AP = BP = CP
That is, (AP)2=(BP)2=(CP)2
This means (AP)2=(BP)

$\Rightarrow(x-5)^{2}+(y-3)^{2}=(x-5)^{2}+(y+5)^{2}$

$\Rightarrow x^{2}-10 x+25+y^{2}-6 y+9=x^{2}-10 x+25+y^{2}+10 y+25$

$\Rightarrow x^{2}-10 x+y^{2}-6 y+34=x^{2}-10 x+y^{2}+10 y+50$

$\Rightarrow x^{2}-10 x+y^{2}-6 y-x^{2}+10 x-y^{2}-10 y=50-34$

$\Rightarrow-16 y=16$

$\Rightarrow y=-\frac{16}{16}=-1$

And $(B P)^{2}=(C P)^{2}$

$\Rightarrow(x-5)^{2}+(y+5)^{2}=(x-1)^{2}+(y+5)^{2}$

$\Rightarrow x^{2}-10 x+25+y^{2}+10 y+25=x^{2}-2 x+1+y^{2}+10 y+25$

$\Rightarrow x^{2}-10 x+y^{2}+10 y+50=x^{2}-2 x+y^{2}+10 y+26$

$\Rightarrow x^{2}-10 x+y^{2}+10 y-x^{2}+2 x-y^{2}-10 y=26-50$

$\Rightarrow-8 x=-24$

$\Rightarrow x=\frac{-24}{-8}=3$

Hence, the required point is (3, −1).