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

Question

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

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Answer

Let the point on $x y$-plane be $P(x, y, 0)$.
Now $P$ is equidistance from $A(1,-1,0), B(2,1,2)$ and $C(3,2,-1)$.
So, AP - BP - CP
Now,
$(A P)^2=(x-1)^2+(y+1)^2+(0-0)^2$
$(B P)^2=(x-2)^2+(y-1)^2+(0-2)^2$
$(C P)^2=(x-3)^2+(y-2)^2+(0+1)^2$
$(A P)^2=(B P)^2 \Rightarrow(x-1)^2+(y+1)^2-(x-2)^2+(y-1)^2+4$
$\Rightarrow x^2+1-2+y^2+1+2 y+z^2-x^2+4-4 x+y^2+1-2 y+4$
$\Rightarrow 2 x+4 y=7 \ldots(i)$
$(B P)^2=(C P)^2 \Rightarrow(x-2)^2+(y-1)^2+4=(x-3)^2+(y-2)^2+1$
$\Rightarrow x^2+4-4 x+y^2+1-2 y+z^2+4=x^2+9-6 x+y^2+4-4 y+1$
$\Rightarrow 2 x+2 y=5 \ldots(\text { ii) }$
$(A P)^2=(C P)^2 \Rightarrow(x-1)^2+(y+1)^2+(x-3)^2+(y-2)^2+1$
$\Rightarrow x^2+1-2 x+y^2+1+2 y=x^2+9-6 x+y^2+4-4 y+1$
$\Rightarrow 4 x+5 y=12 \ldots \text { (iii) }$
Solving equation (i) and (ii) we get
$y=1, x=\frac{3}{2}$
Put x and y in equation (iii)
y = 1, $\text{x} = \frac{3}{2}$
Put x and y in equation (iii)
$4\Big(\frac{3}{2}\Big) + 6(1) = 12$
$12-12$
So, the required point is $\Big(\frac{3}{2},\ 1,\ 0\Big)$

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