Question

If a point P is equidistant from A(1, 3), B(-3, 5) and C(5, -1), then PA is equal to

• A. (A) 5
• B. (B) 5√5
• C. (C) 25
• D. (D) 5√10

Answer 1

Answer: (D)

(D) 5√10

Answer 2

Explanation:
Perpendicular bisector of A(1,3) and B(−3,5) is2x(x1−x2)+2y(y1−y2)=(x12+y12)⇒2x(1+3)+2y(3−5)=(1+9)⇒−(x22+y22)⇒2x−4y=10−34Similarly, perpendicular bisector of A(1,3) and C(5−1) is2x(1−5)+2y(3+1)=(1+9)−(25+1)−8x+8y=10−26−x+y=−2⇒x−y−2=0Point of intersection of Eqs. (i) and(ii), is P=(−8,−10)Then, the distance between P and A isPA=(1+8)2+(3+10)2=81+169=250=510