Skip to content
Question of 72

Q.Find the equation of the set of points which are equidistant from the points (1,2,3)(1, 2, 3) and (3,2,−1)(3, 2, -1).

Rajasthan RbseRajasthan Board Senior Secondary Part-I Examination 2026Subjective· 4mImportance★★★★★
0% · 0/72 Questions
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

The required equation is x−2z=0x-2z=0.

Let P(x,y,z)P(x,y,z) be a point equidistant from A(1,2,3)A(1,2,3) and B(3,2,−1)B(3,2,-1). Then PA=PBPA = PB, so PA2=PB2PA^2=PB^2:

(x−1)2+(y−2)2+(z−3)2=(x−3)2+(y−2)2+(z+1)2(x-1)^2+(y-2)^2+(z-3)^2 = (x-3)^2+(y-2)^2+(z+1)^2

The (y−2)2(y-2)^2 term appears on both sides and cancels:

(x−1)2+(z−3)2=(x−3)2+(z+1)2(x-1)^2+(z-3)^2 = (x-3)^2+(z+1)^2

Expand each square:

x2−2x+1+z2−6z+9=x2−6x+9+z2+2z+1x^2-2x+1 + z^2-6z+9 = x^2-6x+9 + z^2+2z+1

Cancel x2x^2 and z2z^2 from both sides:

−2x+1−6z+9=−6x+9+2z+1-2x+1-6z+9 = -6x+9+2z+1

…

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.