Question

Determine the points in xy-plan are equidistant from the points A(1, –1, 0), B(2, 1, 2) and C(3, 2, –1).

Answer 1

We know that the z-coordinate of every point on the xy-plane is zero.
 So, let P (x, y, 0) be a point on the xy-plane such that PA = PB = PC 
Now, PA = PB

\[\Rightarrow P A^2 = P B^2 \]
\[ \Rightarrow \left( x - 1 \right)^2 + \left( y + 1 \right)^2 + \left( 0 - 0 \right)^2 = \left( x - 2 \right)^2 + \left( y - 1 \right)^2 + \left( 0 - 2 \right)^2 \]
\[ \Rightarrow x^2 - 2x + 1 + y^2 + 2y + 1 = x^2 - 4x + 4 + y^2 - 2y + 1 + 4\]
\[ \Rightarrow - 2x + 4x + 2y + 2y + 2 - 9 = 0\]
\[ \Rightarrow 2x + 4y - 7 = 0\]
\[ \Rightarrow 2x + 4y = 7 . . . (1)\]
\[ PB = PC\]
\[ \Rightarrow P B^2 = P C^2 \]
\[ \Rightarrow \left( x - 2 \right)^2 + \left( y - 1 \right)^2 + \left( 0 - 2 \right)^2 = \left( x - 3 \right)^2 + \left( y - 2 \right)^2 + \left( 0 + 1 \right)^2 \]
\[ \Rightarrow x^2 - 4x + 4 + y^2 - 2y + 1 + 4 = x^2 - 6x + 9 + y^2 - 4y + 4 + 1\]
\[ \Rightarrow - 4x + 6x - 2y + 4y + 9 - 14 = 0\]
\[ \Rightarrow 2x + 2y - 5 = 0\]
\[ \Rightarrow x + y = \frac{5}{2}\]
\[ \Rightarrow x = \frac{5}{2} - y . . . \left( 2 \right)\]
\[\text{ Putting the value of x in equation } \left( 1 \right): \]
\[ 2\left( \frac{5}{2} - y \right) + 4y = 7\]
\[ \Rightarrow 5 - 2y + 4y = 7\]
\[ \Rightarrow 5 + 2y = 7\]
\[ \Rightarrow 2y = 2\]
\[ \Rightarrow y = \frac{2}{2}\]
\[ \therefore y = 1\]
\[\text{ Putting the value of y in equation } \left( 2 \right): \]
\[x = \frac{5}{2} - 1\]
\[x = \frac{5 - 2}{2}\]
\[x = \frac{3}{2}\]
\[\text{ Hence, the required point is } \left( \frac{3}{2}, 1, 0 \right) . \]