3D Geometry
Equidistant point in coordinate plane; triangle classification
nta_pyq_2025_apr
Grade 12
Question:
Let $A(x,y,z)$ be a point in $xy$-plane, which is equidistant from three points $P(0,3,2)$, $Q(2,0,3)$ and $R(0,0,1)$. Let $B=(1,4,-1)$ and $C=(2,0,-2)$. Then among the statements
(S1): $\triangle ABC$ is an isosceles right angled triangle, and
(S2): the area of $\triangle ABC$ is $\dfrac{9\sqrt{2}}{2}$
both are true
only (S2) is true
only (S1) is true
both are false
Step-by-Step Solution
Key Concept: Set $z=0$ and use $AP=AQ=AR$ to find $A$, then compute $AB$, $AC$, $BC$ to verify the isosceles right-angle condition and the area formula.
$z=0$. $AP^2=AQ^2$: $x^2+(y-3)^2+4=( x-2)^2+y^2+9 \Rightarrow x=3$.
$AP^2=AR^2$: $9+(y-3)^2+4=9+y^2+1 \Rightarrow -6y+9+4=1 \Rightarrow y=2$.
$A=(3,2,0)$, $B=(1,4,-1)$, $C=(2,0,-2)$.
$AB=\sqrt{4+4+1}=3$, $AC=\sqrt{1+4+4}=3$, $BC=\sqrt{1+16+1}=\sqrt{18}=3\sqrt{2}$.
$AB=AC=3$ (isosceles) and $AB^2+AC^2=18=BC^2$ (right angle at $A$). S1 is true.
Area $=\tfrac{1}{2}\times3\times3=\tfrac{9}{2}\neq\tfrac{9\sqrt{2}}{2}$. S2 is false.
Only S1 is true.
Correct Answer: 3