Let A(x,y,z) be a point in xy-plane, which is equidistant from three points (0,3,2),(2,0,3…

Question

Let $A(x,y,z)$ be a point in $xy$-plane, which is equidistant from three points $(0,3,2),(2,0,3)$ and $(0,0,1)$.

Let $B=(1,4,-1)$ and $C=(2,0,-2)$. Then among the statements

(S1) : $	riangleABC$ is an isosceles right angled triangle, and

(S2) : the area of $	riangleABC$ is $\frac{9\sqrt{2}}{2}$,

Accepted Answer

$\begin{gathered} A(x,y,z)P(0,3,2),Q(2,0,3),R(0,0,1) \ AP=AQ=AR \ {x}^{2}+(y-3{)}^{2}+(z-2{)}^{2}=(x-2{)}^{2}+{y}^{2}+(z-3{)}^{2}={x}^{2}+ \ {y}^{2}+(z-1{)}^{2} \end{gathered}$

In xy plane $z=0$

So, ${x}^{2}-4x+4+{y}^{2}+9={x}^{2}+{y}^{2}+1$

$\begin{gathered} \Rightarrow y=2 \ x=3 \ 9+{y}^{2}-6y+9+4={x}^{2}+{y}^{2}+1 \end{gathered}$

So, $A(3,2,0)$ also $B(1,4,-1)\&C(2,0,-2)$

Now $AB=\sqrt{4+4+1}=3$

$\begin{gathered} AC=\sqrt{1+4+4}=3 \ BC=\sqrt{1+16+1}=\sqrt{18} \ AB=AC \ \Delta{AB}^{2}+{AC}^{2}={BC}^{2} \ \Delta \ 	riangleABC=\frac{1}{2}	imes \ \frac{1}{2}	imes 3	imes 3=\frac{9}{2} \ {S}_{1} \end{gathered}$

Solutions · 1