Determine the points in (i) xy-plane (ii) yz-plane and (iii) zx-plane which are equidistant from the points A(1, -1, 0), B(2, 1, 2) and C(3, 2, -1).
(i) xy-plane
Given: Points A(1, -1, 0), B(2, 1, 2) and C(3, 2, -1)
To find: the point on xy-plane which is equidistant from the points
As we know z = 0 in xy-plane.
Let P(x, y, 0) any point in xy-plane
According to the question:
PA = PB = PC
⇒ PA2 = PB2 = PC2
Formula used:
The distance between any two points (a, b, c) and (m, n, o) is given by,
Therefore,
Distance between P(x, y, 0) and A(1, -1, 0) is PA,
The distance between P(x, y, 0) and B(2, 1, 2) is PB,
Distance between P(x, y, 0) and C(3, 2, -1) is PC,
As PA2 = PB2
(x – 1)2+ (y + 1)2 = (x – 2)2 + (y – 1)2 + 4
⇒ x2+ 1 – 2x + y2 + 1 + 2y = x2+ 4 – 4x + y2 + 1 – 2y + 4
⇒ – 2x + 2 + 2y = 9 – 4x – 2y
⇒ – 2x + 2 + 2y – 9 + 4x + 2y = 0
⇒ 2x + 4y – 7 = 0
⇒ 2x = - 4y + 7……………………(1)
As PA2 = PC2
(x – 1)2+ (y + 1)2 = (x – 3)2 + (y – 2)2 + 1
⇒ x2+ 1 – 2x + y2 + 1 + 2y = x2+ 9 – 6x + y2 + 4 – 4y + 1
⇒ – 2x + 2 + 2y = 14 – 6x – 4y
⇒ – 2x + 2 + 2y – 14 + 6x + 4y = 0
⇒ 4x + 6y – 12 = 0
⇒ 2(2x + 3y – 6) = 0
Put the value of 2x from (1):
⇒ 7 – 4y + 3y – 6 = 0
⇒ – y + 1 = 0
⇒ y = 1
Put this value of y in (1):
2x = 7 – 4y
⇒ 2x = 7 – 4(1)
⇒ 2x = 3
Hence point in xy-plane is equidistant from A, B and C
(ii) yz-plane
Given: Points A(1, -1, 0), B(2, 1, 2) and C(3, 2, -1)
To find: the point on yz-plane which is equidistant from the points
As we know x = 0 in yz-plane.
Let Q(0, y, z) any point in yz-plane
According to the question:
QA = QB = QC
⇒ QA2 = QB2 = QC2
Formula used:
The distance between any two points (a, b, c) and (m, n, o) is given by,
Therefore,
Distance between Q(0, y, z) and A(1, -1, 0) is QA,
The distance between Q(0, y, Z) and B(2, 1, 2) is QB,
Distance between Q(0, y, z) and C(3, 2, -1) is QC,
As QA2 = QB2
1 + z2+ (y + 1)2 = (z – 2)2 + (y – 1)2 + 4
⇒ z2+ 1 + y2 + 1 + 2y = z2+ 4 – 4z + y2 + 1 – 2y + 4
⇒ 2 + 2y = 9 – 4z – 2y
⇒ 2 + 2y – 9 + 4z + 2y = 0
⇒ 4y + 4z – 7 = 0
⇒ 4z = –4y + 7
As QA2 = QC2
1 + z2+ (y + 1)2 = (z + 1)2 + (y – 2)2 + 9
⇒ z2+ 1 + y2 + 1 + 2y = z2+ 1 + 2z + y2 + 4 – 4y + 9
⇒ 2 + 2y = 14 + 2z – 4y
⇒ 2 + 2y – 14 – 2z + 4y = 0
⇒ –2z + 6y – 12 = 0
⇒ 2(–z + 3y – 6) = 0
Put the value of z from (1):
⇒ 12y + 4y – 7 – 24 = 0
⇒ 16y – 31 = 0
Put this value of y in (1):
Hence point in yz-plane is equidistant from A, B and C
(iii) xz-plane
Given: Points A(1, -1, 0), B(2, 1, 2) and C(3, 2, -1)
To find: the point on xz-plane which is equidistant from the points
As we know y = 0 in xz-plane.
Let R(x, 0, z) any point in xz-plane
According to the question:
RA = RB = RC
⇒ RA2 = RB2 = RC2
Formula used:
The distance between any two points (a, b, c) and (m, n, o) is given by,
Therefore,
Distance between R(x, 0, z) and A(1, -1, 0) is RA,
Distance between R(x, 0, z) and B(2, 1, 2) is RB,
Distance between R(x, 0, z) and C(3, 2, -1) is RC,
As RA2 = RB2
1 + z2+ (x – 1)2 = (z – 2)2 + (x – 2)2 + 1
⇒ z2+ 1 + x2 + 1 – 2x = z2+ 4 – 4z + x2 + 4 – 4x + 1
⇒ 2 – 2x = 9 – 4z – 4x
⇒ 2 + 4z – 9 + 4x – 2x = 0
⇒ 2x + 4z – 7 = 0
⇒ 2x = –4z + 7……………………………(1)
As RA2 = RC2
1 + z2+ (x – 1)2 = (z + 1)2 + (x – 3)2 + 4
⇒ z2+ 1 + x2 + 1 – 2x = z2+ 1 + 2z + x2 + 9 – 6x + 4
⇒ 2 – 2x = 14 + 2z – 6x
⇒ 2 – 2x – 14 – 2z + 6x = 0
⇒ –2z + 4x – 12 = 0
⇒ 2(2x) = 12 + 2z
Put the value of 2x from (1):
⇒ 2(–4z + 7) = 12 + 2z
⇒ –8z + 14 = 12 + 2z
⇒ 14 – 12 = 8z + 2z
⇒ 10z = 2
Put this value of z in (1):
Hence point in xz-plane is equidistant from A, B and C