Questions
The point in plane which is equidistant from the points and
a
(3,1,0)
b
−32,1,0
c
32,1,0
d
3110,0,15
detailed solution
Correct option is C
Suppose that P(x1,y1,0) be any point in the xy- plane and suppose that it is equidistant from the points A(1,-1,0),B(2,1,2)and C(3,2,-1)it gives PA=PB=PC⇒(x−1)2+(y+1)2+0=(x−2)2+(y−1)2+4⇒2x+4y=7and ⇒(x−1)2+(y+1)2+2=(x−3)2+(y−2)2+1⇒2x+3y=6Solving the above equations y=1.x=32Therefore, the required point is 32,1,0Talk to our academic expert!
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