Question
Determine the points xy-plane equidistant from the points A(1, -1, 0), B(2, 1, 2) and C(3, 2, -1).
Let the point on xy-plane be P(x, y, 0).
Now P is equidistance from A(1, -1, 0), B(2, 1, 2) and C(3, 2, -1).
So, AP - BP - CP
Now,
(AP)2 = (x - 1)2 + (y + 1)2 + (0 - 0)2
(BP)2 = (x - 2)2 + (y - 1)2 + (0 - 2)2
(CP)2 = (x - 3)2 + (y - 2)2 + (0 + 1)2
(AP)2 = (BP)2 ⇒ (x - 1)2 + (y + 1)2 - (x - 2)2 + (y - 1)2 + 4
⇒ x2 + 1 - 2 + y2 + 1 + 2y + z2 - x2 + 4 - 4x + y2 + 1 - 2y + 4
⇒ 2x + 4y = 7 ... (i)
(BP)2 = (CP)2 ⇒ (x - 2)2 + (y - 1)2 + 4 = (x - 3)2 + (y - 2)2 + 1
⇒ x2 + 4 - 4x + y2 + 1 - 2y + z2 + 4 = x2 + 9 - 6x + y2 + 4 - 4y +1
⇒ 2x + 2y = 5 ... (ii)
(AP)2 = (CP)2 ⇒ (x - 1)2 + (y + 1)2 + (x - 3)2 + (y - 2)2 + 1
⇒ x2 + 1 - 2x + y2 + 1 + 2y = x2 + 9 - 6x + y2 + 4 - 4y + 1
⇒ 4x + 5y = 12 ... (iii)
Solving equation (i) and (ii) we get
y = 1, $\text{x} = \frac{3}{2}$
Put x and y in equation (iii)
$4\Big(\frac{3}{2}\Big) + 6(1) = 12$
$12-12$
So, the required point is $\Big(\frac{3}{2},\ 1,\ 0\Big)$
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