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

Question

Let $A(x,y,z)$ be a point in the $xy$-plane which is equidistant from the 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): $	riangle ABC$ is an isosceles right-angled triangle, and

(S2): the area of $	riangle ABC$ is $\dfrac{9\sqrt2}{2}$,

Accepted Answer

$$\begin{aligned}
& A(x, y, z) \text { Let } P(0,3,2), Q(2,0,3), R(0,0,1) \
& A P=A Q=A R \
& x^2+(y-3)^2+(z-2)^2=(x-2)^2+y^2+(z-3)^2=x^2+ \
& y^2+(z-1)^2
\end{aligned}$$

In xy plane $\mathrm{z}=0$

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

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

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

Now $A B=\sqrt{4+4+1}=3$

$$\begin{aligned}
&\begin{aligned}
& \mathrm{AC}=\sqrt{1+4+4}=3 \
& \mathrm{BC}=\sqrt{1+16+1}=\sqrt{18} \
& \mathrm{AB}=\mathrm{AC} \
& \text { isosceles } \Delta & \mathrm{AB}^2+\mathrm{AC}^2=\mathrm{BC}^2 \
& \text { right angle } \Delta \
& \text { Area of } \triangle \mathrm{ABC}=\frac{1}{2} \times \text { base.height } \
& \frac{1}{2} \times 3 \times 3=\frac{9}{2}
\end{aligned}\
\&\text { So only } S_1 \text { is true }
\end{aligned}$$

Solutions · 1