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Q.Find the point of z-axis which is equidistant from the points (1, 5, 7) and (5, 1, -4).

(a) (0, 0, 3/2)
(b) (0, 0, 5)
(c) (0, 5, 0)
(d) (4, 2, 3).
West Bengal WbchseWest Bengal HS First Year (WBCHSE Class XI) Annual Examination 2018MCQ· 1mImportance★★★★★est
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Setting the point (0,0,z)(0,0,z) equidistant from (1,5,7)(1,5,7) and (5,1,−4)(5,1,-4) and solving gives z=3/2z=3/2.

A general point on the zz-axis is P=(0,0,z)P=(0,0,z). Using the 3D distance formula, equate the squared distances from PP to A=(1,5,7)A=(1,5,7) and B=(5,1,−4)B=(5,1,-4):

PA2=(0−1)2+(0−5)2+(z−7)2=1+25+(z−7)2=26+(z−7)2PA^2=(0-1)^2+(0-5)^2+(z-7)^2=1+25+(z-7)^2=26+(z-7)^2

PB2=(0−5)2+(0−1)2+(z+4)2=25+1+(z+4)2=26+(z+4)2PB^2=(0-5)^2+(0-1)^2+(z+4)^2=25+1+(z+4)^2=26+(z+4)^2 …

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