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JEE Main 202528 JanuaryMorning Shift
Exam
JEE Main
Year
2025
Date
28 January
Shift
Morning Shift
Subject
Mathematics
Chapter
Three Dimensional Geometry

JEE Main 2025 · 28 January · Morning Shift · Mathematics · Three Dimensional Geometry

Let A(x, y, z) be a point in x y -plane, which is equidistant from...

Let A(x,y,z)\mathrm{A}(x, y, z) be a point in xyx y-plane, which is equidistant from three points (0,3,2),(2,0,3)(0,3,2),(2,0,3) and ( 0,0,10,0,1 ). Let B=(1,4,−1)\mathrm{B}=(1,4,-1) and C=(2,0,−2)\mathrm{C}=(2,0,-2). Then among the statements (S1) : △ABC\triangle \mathrm{ABC} is an isosceles right angled triangle, and (S2) : the area of △ABC\triangle \mathrm{ABC} is 922\frac{9 \sqrt{2}}{2},

Options

  1. Aboth are true
  2. Bonly (S2) is true
  3. Conly (S1) is true
  4. Dboth are false

Correct answer

C. only (S1) is true

Step-by-step solution

A(x,y,z) Let P(0,3,2),Q(2,0,3),R(0,0,1)AP=AQ=ARx2+(y−3)2+(z−2)2=(x−2)2+y2+(z−3)2=x2+y2+(z−1)2\begin{aligned} & \mathrm{A}(\mathrm{x}, \mathrm{y}, \mathrm{z}) \text{ Let } \mathrm{P}(0,3,2), \mathrm{Q}(2,0,3), \mathrm{R}(0,0,1) \\ & \mathrm{AP}=\mathrm{AQ}=\mathrm{AR} \\ & \mathrm{x}^2+(\mathrm{y}-3)^2+(\mathrm{z}-2)^2=(\mathrm{x}-2)^2+\mathrm{y}^2+(\mathrm{z}-3)^2=\mathrm{x}^2+ \\ & \mathrm{y}^2+(\mathrm{z}-1)^2 \end{aligned} In x y plane z=0z=0 So, x2−4x+4+y2+9=x2+y2+1x^2-4 x+4+y^2+9=x^2+y^2+1x=39+y2−6y+9+4=x2+y2+1\begin{aligned} & x=3 \\ & 9+y^2-6 y+9+4=x^2+y^2+1 \end{aligned}So, $\mathrm{A}(3,2,0)$ also $\mathrm{B}(1,4,-1) \& \mathrm{C}(2,0,-2)$ Now $A B=\sqrt{4+4+1}=3\begin{aligned} & \mathrm{AC}=\sqrt{1+4+4}=3 \\ & \mathrm{BC}=\sqrt{1+16+1}=\sqrt{18} \end{aligned}$AB=AC\mathrm{AB}=\mathrm{AC} isosceles Δ&AB2+AC2=BC2\Delta \& \mathrm{AB}^2+\mathrm{AC}^2=\mathrm{BC}^2 right angle Δ\Delta Area of △ABC=12×\triangle \mathrm{ABC}=\frac{1}{2} \times base.height 12×3×3=92\frac{1}{2} \times 3 \times 3=\frac{9}{2} So only S1S_1 is true
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