The point P is equidistant from A(1,3) , B(-3,5) and C(5,-1) , then P A is equal to :

Step-by-step solution

Let co-ordinates of P are (x, y) . Since, P is equidistant from A, B, C , then P A^2=P B^2 ...(i) and P B^2=P C^2 ...(ii) From Eq. (i), (x-1)^2+(y-3)^2=(x+3)^2+(y-5)^2 x^2+1-2 x+y^2+9-6 y=x^2+9+6 x+y^2+25-10 y 8 x-4 y+24=0 2 x-y+6=0 ...(iii) From Eq. (ii), (x+3)^2+(y-5)^2=(x-5)^2+(y+1)^2 x^2+9+6 x+y^2+25-10 y=x^2+25-10 x+y^2+1+2 y 16 x-12 y+8=0 4 x-3 y+2=0 ...(iv) On solving Eqs. (iii) and (iv), we get x=-8, y=-10 Now, P A^2=(-8-1)^2+(-10-3)^2=81+169=250P A= 250 =5 10