Question

Determine the points in (i) xy-plane (ii) yz-plane and (iii) zx-plane which are equidistant from the points A(1, –1, 0), B(2, 1, 2) and C(3, 2, –1).

Answer 1

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

(ii) We know that the x-coordinate of every point on the yz-plane is zero.
So, let P (0, y, z) be a point on the yz-plane such that PA = PB = PC
Now, PA = PB
$\Rightarrow {\left(0-1\right)}^2+{\left(y+1\right)}^2+{\left(z-0\right)}^2={\left(0-2\right)}^2+{\left(y-1\right)}^2+{\left(z-2\right)}^2$
$$\begin{gathered} \Rightarrow 1+y^2+2y+1+z^2=4+y^2-2y+1+z^2-4z+4 \\ \Rightarrow 2y+2=-2y-4z+9 \\ \Rightarrow 2y+2y-4z=9-2 \\ \Rightarrow 4y-4z=7 \\ \Rightarrow y-z=\frac{7}{4}...\left(1\right) \\ PB=PC \\ \Rightarrow P{B}^2=P{C}^2 \\ \Rightarrow {\left(0-2\right)}^2+{\left(y-1\right)}^2+{\left(z-2\right)}^2={\left(0-3\right)}^2+{\left(y-2\right)}^2+{\left(z+1\right)}^2 \\ \Rightarrow 4+y^2-2y+1+z^2-4z+4=9+y^2-4y+4+z^2+2z+1 \\ \Rightarrow -2y-4z+9=-4y+2z+14 \\ \Rightarrow -2y+4y-4z-2z=14-9 \\ \Rightarrow 2y-6z=5 \\ \Rightarrow y-3z=\frac{5}{2} \\ \therefore y=\frac{5}{2}+3z\left(2\right) \\ \mathrm{Putting}\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{y}\mathrm{in}\mathrm{equation}\left(1\right): \\ y-z=\frac{7}{4} \\ \Rightarrow \frac{5}{2}+3z-z=\frac{7}{4} \\ \Rightarrow 2z=\frac{7}{4}-\frac{5}{2} \\ \Rightarrow 2z=\frac{7-10}{4} \\ \Rightarrow 2z=\frac{-3}{4} \\ \therefore z=\frac{-3}{8} \\ \mathrm{Putting}\mathrm{the}\mathrm{value}\mathrm{of}z\mathrm{in}\mathrm{equation}\left(2\right): \\ y=\frac{5}{2}+3z \\ \Rightarrow y=\frac{5}{2}+3\left(\frac{-3}{8}\right) \\ \Rightarrow y=\frac{5}{2}-\frac{9}{8} \\ \Rightarrow y=\frac{20-9}{8} \\ \therefore y=\frac{11}{8} \end{gathered}$$
Hence, the required point is $\left(0,\frac{11}{8},\frac{-3}{8}\right)$.

(iii) 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
$\Rightarrow {\left(x-1\right)}^2+{\left(0+1\right)}^2+{\left(z-0\right)}^2={\left(x-2\right)}^2+{\left(0-1\right)}^2+{\left(x-2\right)}^2$
$$\begin{gathered} \Rightarrow x^2+1-2x+1+z^2=x^2-4x+4+1+z^2-4z+4 \\ \Rightarrow -2x+2=-4x-4z+9 \\ \Rightarrow -2x+4x-4z=7 \\ \Rightarrow 2x-4z=7 \\ \Rightarrow x-2z=\frac{7}{2}...\left(1\right) \\ PB=PC \\ \Rightarrow P{B}^2=P{C}^2 \\ \Rightarrow {\left(x-2\right)}^2+{\left(0-1\right)}^2+{\left(z-2\right)}^2={\left(x-3\right)}^2+{\left(0-2\right)}^2+{\left(z+1\right)}^2 \\ \Rightarrow x^2-4x+4+1+z^2-4z+4=x^2-6x+9+4+z^2+2z+1 \\ \Rightarrow -4x-4z+9=-6x+2z+14 \\ \Rightarrow -4x+6x-4z-2z=14-9 \\ \Rightarrow 2x-6z=5 \\ \Rightarrow x-3z=\frac{5}{2} \\ \therefore x=\frac{5}{2}+3z...\left(2\right) \\ \mathrm{Putting}\mathrm{the}\mathrm{value}\mathrm{of}\mathit{}x\mathit{}\mathrm{in}\mathrm{equation}\left(1\right): \\ x-2z=\frac{7}{2} \\ \Rightarrow \frac{5}{2}+3z-2z=\frac{7}{2} \\ \Rightarrow \frac{5}{2}+z=\frac{7}{2} \\ \Rightarrow z=\frac{7}{2}-\frac{5}{2} \\ \Rightarrow z=\frac{7-5}{2} \\ \Rightarrow z=\frac{2}{2} \\ \therefore z=1 \\ \mathrm{Putting}\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{z}\mathrm{in}\mathrm{equation}\left(2\right): \\ x=\frac{5}{2}+3z \\ \Rightarrow x=\frac{5}{2}+3\left(1\right) \\ \Rightarrow x=\frac{5}{2}+3 \\ \Rightarrow x=\frac{5+6}{2} \\ \therefore x=\frac{11}{2} \end{gathered}$$
Hence, the required point is $\left(\frac{11}{2},0,1\right)$.