Q.
The point on z-axis which is equidistant from the points (1,5,7) and (5,1,-4)
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a
0,0,32
b
0,0,1
c
0,0,132
d
0,0,5
answer is A.
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Detailed Solution
Suppose that P(0,0,z) is any point on z- axis such that it is equidistant from the points A(1,5,7)and B(5,1,-4) It gives PA=PB(z−7)2=(z+4)2−14z+49=8z+16 22z=33 z=32Therefore, the point is P0,0,32
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