Question:
12. Determine the point in XY plane which is equidistant from the points A (1, –1, 0) B(2, 1, 2) and C(3, 2, –1).
Answer:
Let the point in xy-plane is (x, y, 0)
Given points are: A(1, –1, 0), B(2, 1, 2), C(3, 2, –1)
Now, PA = PB = PC
Now, PA2 = PB2
=> (x - 1)2 + (y + 1)2 = (x - 2)2 + (y - 1)2 + (0 - 2)2
=> x2 + 1 - 2x + y2 + 1 + 2y = x2 + 4 - 4x + y2 + 1 - 2y + 4
=> -2x + 2y + 2 = -4x - 2y + 9
=> 2x + 4y = 7 .............1
Again, PB2 = PC2
=> (x - 2)2 + (y - 1)2 + + (0 - 2)2 = (x - 3)2 + (y - 2)2 + (0 + 1)2
=> x2 + 4 - 4x + y2 + 1 - 2y + 4 = x2 + 9 - 6x + y2 + 4 - 4y + 1
=> -4x - 2y + 9 = -6x - 4y + 14
=> 2x + 2y = 5 .............2
After solving equation 1 and 2, we get
x = 3/2, y = 1
So the points is (3/2, 1, 0)
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