pa school requirements

Find the co-ordinates of a point which is equidistant from the points (-2,9), (10,-7) and (12,-5) Solution Let the co-ordinates of the required point P be (x,y) which is equidistant from the given points A(-2,9),B(10,-7) and C(12,-5) Now; $PA=\sqrt{(x+2)^{2}+(y-9)^{2}}$ $PB=\sqrt{(x-10)^{2}+(y+7)^{2}}$ $PC=\sqrt{(x-12)^{2}+(y+5)^{2}}$ Since PA=PB $\therefore PA^{2}=PB^{2}$ $\Rightarrow(x+2)^{2}+(y-9)^{2}=(x-10)^{2}+(y+7)^{2}$ $\Rightarrow x^{2}+4x+4+y^{2}-18y+81=x^{2}-20x+100+y^{2}+14y+49$ $\Rightarrow24x-32y=64$ $\Rightarrow3x-4y=8$.......(i) Also, since PA=PC $PA^{2}=PC^{2}$ $\Rightarrow(x+2)^{2}+(y-9)^{2}=(x-12)^{2}+(y+5)^{2}$ $\Rightarrow x^{2}+4x+4+y^{2}-18y+81=x^{2}-24x+144+y^{2}+10y+25$ $\Rightarrow28x-28y=84$ $\Rightarrow x-y=3$....(ii) Solving equation (i) & (ii); x=4;y=1 Hence the co-ordinates of the required point P are (4,1) Ans.