Question

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

Answer 1

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

 Now, PA = PB

\[\Rightarrow \left( x - 1 \right)^2 + \left( 0 + 1 \right)^2 + \left( z - 0 \right)^2 = \left( x - 2 \right)^2 + \left( 0 - 1 \right)^2 + \left( x - 2 \right)^2\]

\[\Rightarrow x^2 + 1 - 2x + 1 + z^2 = x^2 - 4x + 4 + 1 + z^2 - 4z + 4\]
\[ \Rightarrow - 2x + 2 = - 4x - 4z + 9\]
\[ \Rightarrow - 2x + 4x - 4z = 7\]
\[ \Rightarrow 2x - 4z = 7\]
\[ \Rightarrow x - 2z = \frac{7}{2} . . . \left( 1 \right)\]
\[ PB = PC\]
\[ \Rightarrow P B^2 = P C^2 \]
\[ \Rightarrow \left( x - 2 \right)^2 + \left( 0 - 1 \right)^2 + \left( z - 2 \right)^2 = \left( x - 3 \right)^2 + \left( 0 - 2 \right)^2 + \left( z + 1 \right)^2 \]
\[ \Rightarrow x^2 - 4x + 4 + 1 + z^2 - 4z + 4 = x^2 - 6x + 9 + 4 + z^2 + 2z + 1\]
\[ \Rightarrow - 4x - 4z + 9 = - 6x + 2z + 14\]
\[ \Rightarrow - 4x + 6x - 4z - 2z = 14 - 9\]
\[ \Rightarrow 2x - 6z = 5\]
\[ \Rightarrow x - 3z = \frac{5}{2}\]
\[ \therefore x = \frac{5}{2} + 3z . . . \left( 2 \right)\]
\[\text{ Putting the value of x in equation } \left( 1 \right): \]
\[ x - 2z = \frac{7}{2}\]
\[ \Rightarrow \frac{5}{2} + 3z - 2z = \frac{7}{2}\]
\[ \Rightarrow \frac{5}{2} + z = \frac{7}{2}\]
\[ \Rightarrow z = \frac{7}{2} - \frac{5}{2}\]
\[ \Rightarrow z = \frac{7 - 5}{2}\]
\[ \Rightarrow z = \frac{2}{2}\]
\[ \therefore z = 1\]
\[\text{ Putting the value of z in equation } \left( 2 \right): \]
\[ x = \frac{5}{2} + 3z\]
\[ \Rightarrow x = \frac{5}{2} + 3\left( 1 \right)\]
\[ \Rightarrow x = \frac{5}{2} + 3\]
\[ \Rightarrow x = \frac{5 + 6}{2}\]
\[ \therefore x = \frac{11}{2}\]

Hence, the required point is \[\left( \frac{11}{2}, 0, 1 \right)\]