A) \[5\sqrt{5}\]
B) \[5\]
C) \[5\sqrt{10}\]
D) \[25\]
Correct Answer: C
Solution :
Given points are \[A(1,3),B(-3,5)\]and\[C(5,-1)\]Mid-piont of AB is\[(-1,4),\]slope of\[AB=-\frac{1}{2}\]. Perpendicular bisector of AB,\[y-4=2(x+1)\] \[\Rightarrow \] \[2x-y+6=0\] ...(i) Similarly, perpendicular bisector of AC is \[x-y-2=0\] ...(ii) Point of intersection of Eqs. (i) and (ii) is \[P=(-8,-10)\] Then, \[PA=\sqrt{{{(1+8)}^{2}}+{{(3+10)}^{2}}}\] \[=\sqrt{81+169}=5\sqrt{10}\]You need to login to perform this action.
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