Skip to content
Exercise 11.2 · Q4

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).

Karnataka PUCTextbookSubjective· 3mImportance★★★★★est
11% · 8/72 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 set of points equidistant from two fixed points in 3D is the perpendicular bisector plane of the segment joining them. For (1,2,3)(1,2,3) and (3,2,−1)(3,2,-1), the equation is x−2z=0x - 2z = 0.

Why this works — the 3D coordinate octants and the idea of a locus

In three-dimensional space, the condition “equidistant from two given points” defines a plane. Why? Because in any geometry (2D or 3D), the set of points at equal distance from two fixed points is the perpendicular bisector of the segment joining them. In 2D, that’s a line; in 3D, it’s a plane — the plane that cuts the segment at its midpoint and stands perpendicular to it.

Think of it this way: if you have two points AA and BB, any point PP that is equally far from AA and BB must lie on the “middle surface” that is orthogonal to ABAB and passes through its midpoint. That surface is flat — a plane.

So our job is to find the equation of that plane. We don’t need to visualise octants here; we just use the distance formula and simplify.


Step-by-step solution

1. Write the distance condition algebraically

Let P(x,y,z)P(x, y, z) be any point satisfying the condition. The distance from PP to A(1,2,3)A(1,2,3) equals the distance from PP to B(3,2,−1)B(3,2,-1).

Using the 3D distance formula:

(x−1)2+(y−2)2+(z−3)2=(x−3)2+(y−2)2+(z+1)2\sqrt{(x-1)^2 + (y-2)^2 + (z-3)^2} = \sqrt{(x-3)^2 + (y-2)^2 + (z+1)^2}

2. Square both sides to remove the square roots

Since both sides are non-negative, squaring is safe:

(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

Notice that (y−2)2(y-2)^2 appears on both sides — they cancel immediately.

3. Expand the remaining squares

Left side: (x−1)2+(z−3)2=(x2−2x+1)+(z2−6z+9)(x-1)^2 + (z-3)^2 = (x^2 - 2x + 1) + (z^2 - 6z + 9)

Right side: (x−3)2+(z+1)2=(x2−6x+9)+(z2+2z+1)(x-3)^2 + (z+1)^2 = (x^2 - 6x + 9) + (z^2 + 2z + 1)

4. Cancel common terms

Both sides have x2x^2 and z2z^2. Subtract them from both sides:

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

Simplify constants: 1+9=101+9 = 10 on left, 9+1=109+1 = 10 on right. So:

−2x−6z+10=−6x+2z+10-2x - 6z + 10 = -6x + 2z + 10

Cancel 1010 from both sides:

−2x−6z=−6x+2z-2x - 6z = -6x + 2z

5. Collect terms to one side …

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.