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+ y^{2} -4y + 4 + z^{2} + 2z + 1 = 1+ y^{2} +2y + 1+ z^{2}

-6y + 2z + 12 = 0 ….(1)

Also, AD = CD

Squaring both sides,

=

9+ y^{2} -4y + 4 + z^{2} + 2z + 1 = 4+ y^{2} - 2y + 1+ z^{2} – 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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