Determine the points zx-plane equidistant from the points A(1, -1… - Vidyadip

Question

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

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Answer

Let $R(x, 0, z)$ be the required point.
So
$(A R)^2=(B R)^2 \Rightarrow(1-x)^2+(-1-0)^2+(0-z)^2=(2-x)+(1-0)^2+(2-z)^2$
$\Rightarrow 1+x^2-2 x+1+z^2=4+x^2-4 x+1+z^2+4 z$
$\Rightarrow 2 x+4 z=7 \ldots(\mathrm{i})$
$(B R)^2=(C R)^2 \Rightarrow(z-z)^2+(1-0)^2+(2-z)^2=(3-x)^2+(2-0)^2(-1-z)^2$
$\Rightarrow 4+x^2-4 x+4+z^2-4 z=9+x^2-6 x+4+1+z^2+2 z$
$\Rightarrow 2 x-6 z=5 \ldots(\text { ii) }$
$(A R)^2=(C R)^2 \Rightarrow(1-x)^2+(1-0)^2+(0-z)^2=(3-x)^2+(2-0)^2+(-1-z)^2$
$\Rightarrow 1+x^2-2 x+1+z^2=9+6 x+4+1+z^2$
$\Rightarrow 4 x-2 z=12 \ldots \text { (iii) }$
Solving equation (i) and (ii) we get
$\text{z}=\frac{1}{5},\ \text{x}=\frac{31}{10}$
Put the value of x and z in equation (iii)
$4x - 2z = 12$
$4\Big(\frac{31}{10}\Big)-2\Big(\frac{1}{5}\Big) = 12$
$\frac{124}{10}+\frac{2}{10}=12$
$\frac{120}{10}=12$
$12=12$
LHS = RHS.
so,
$\text{x}=\frac{31}{10},\ \text{z}=\frac{1}{5}$
Required point $=\Big(\frac{31}{10},\ 0,\ \frac{1}{5}\Big)$

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