Q. 155.0( 1 Vote )

Find the point in yz-plane which is equidistant from the points A(3, 2, -1), B(1, -1, 0) and C(2, 1, 2).

Answer :

The general point on yz plane is D(0, y, z). Consider this point is equidistant to the points A(3, 2, -1), B(1, -1, 0) and C(2, 1, 2).


AD = BD



Squaring both sides,


=


9+ y2 -4y + 4 + z2 + 2z + 1 = 1+ y2 +2y + 1+ z2


-6y + 2z + 12 = 0 ….(1)


Also, AD = CD



Squaring both sides,


=


9+ y2 -4y + 4 + z2 + 2z + 1 = 4+ y2 - 2y + 1+ z2 – 4z + 4


-2y + 6z + 5 = 0 ….(2)


Simultaneously solving equation (1) and (2) we get


Y= 31/16, z = -3/16


The point which is equidistant to the points A(3, 2, -1), B(1, -1, 0) and C(2, 1, 2) is (0, 31/16, -3/16).


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