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

Question

Let $A(x, y, z)$ be a point in $x y$-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) : $	riangle A B C$ is an isosceles right angled triangle, and\ (S2) : the area of $	riangle A B C$ is $\frac{9 \sqrt{2}}{2}$.\

Professor PC Thomas Classes · Answer

(a) : Given, $A(x, y, z)$ be a point in $x y$ - plane. Let the point $P(0,3,2), Q(2,0,3)$ and $R(0,0,1)$\ The distance of the point $A P=A Q=A R$\ $\Rightarrow \quad(x-0)^{2}+(y-3)^{2}+(z-2)^{2}=(x-2)^{2}+ (y-0)^{2}+(z-3)^{2}=(x-0)^{2}+(y-0)^{2}+(z-1)^{2}$\ In $x y$ - plane, $z=0$\ So, $x^{2}-4 x+y^{2}+9+4=x^{2}+y^{2}+1$\ $\Rightarrow-4 x+13=1 \Rightarrow x=3$\ And, $x^{2}+y^{2}-6 y+9+4=x^{2}+y^{2}+1$\ $\Rightarrow \quad y=2$\ So, $A(3,2,0), B(1,4,-1)$ and $C(2,0,-2)$.\ In $	riangle A B C$\ $A B=\sqrt{1+4+4}=3, B C=\sqrt{1+16+1}=\sqrt{18}$, $$ C A=\sqrt{1+4+4}=3 $$ So, $A B=A C$ and $A B^{2}+A C^{2}=(B C)^{2}$\ $	herefore 	riangle A B C$ is an isosceles right angled triangle.\ So, (S1) is true.\ Also, Area of $	riangle A B C=\frac{1}{2} 	imes$ (Base)(Height) $=\frac{1}{2} 	imes 3 	imes 3=\frac{9}{2}$\ So, (S2) is false.\