Question 8 of 13intermediate🔧 ApplyLong Answer4 marks

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).

Correct Answer

Let P(x,y,z)P(x, y, z) be any point equidistant from the points A(1,2,3)A(1, 2, 3) and B(3,2,−1)B(3, 2, -1).

Since PP is equidistant from AA and BB, we have PA = PB.

Using the distance formula between two points:

PA2=(x−1)2+(y−2)2+(z−3)2PA^2 = (x - 1)^2 + (y - 2)^2 + (z - 3)^2 PB2=(x−3)2+(y−2)2+(z+1)2PB^2 = (x - 3)^2 + (y - 2)^2 + (z + 1)^2

Since PA=PBPA = PB, squaring both sides gives PA2=PB2PA^2 = PB^2.

Therefore:

(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

Cancelling (y−2)2(y - 2)^2 from both sides:

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

Expanding both sides:

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

Simplifying:

−2x+1−6z+9=−6x+9+2z+1-2x + 1 - 6z + 9 = -6x + 9 + 2z + 1 −2x−6z+10=−6x+2z+10-2x - 6z + 10 = -6x + 2z + 10 −2x−6z=−6x+2z-2x - 6z = -6x + 2z 4x=8z4x = 8z ∴x = 2z\therefore \quad \textbf{x = 2z}

or x - 2z = 0

This is the required equation of the set of points equidistant from (1,2,3)(1, 2, 3) and (3,2,−1)(3, 2, -1).

Exercise: EXERCISE 11.2 | Q: 4 | (Page Page 6)
For More Understanding

Explanation

This question tests the application of the distance formula in 3D space. The set of points equidistant from two given points forms a plane (the perpendicular bisector plane of the line segment joining the two points).

The student needs to set up the equation using PA=PBPA = PB, apply the distance formula, and simplify algebraically. The (y−2)2(y - 2)^2 terms cancel out since both given points have the same yy-coordinate.

Solution Steps

  1. Step 1: Let P(x,y,z)P(x, y, z) be equidistant from A(1,2,3)A(1, 2, 3) and B(3,2,−1)B(3, 2, -1), so PA=PBPA = PB

  2. Step 2: Apply distance formula: PA2=(x−1)2+(y−2)2+(z−3)2PA^2 = (x-1)^2 + (y-2)^2 + (z-3)^2 and PB2=(x−3)2+(y−2)2+(z+1)2PB^2 = (x-3)^2 + (y-2)^2 + (z+1)^2

  3. Step 3: Set PA2=PB2PA^2 = PB^2 and cancel (y−2)2(y-2)^2 from both sides

  4. Step 4: Expand and simplify: (x−1)2+(z−3)2=(x−3)2+(z+1)2(x-1)^2 + (z-3)^2 = (x-3)^2 + (z+1)^2

  5. Step 5: Solve to get x=2zx = 2z or x−2z=0x - 2z = 0

PreviousNext