We know z = 0 in xy-plane.
We know PA2 = PB2
So, (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)
Since, PA2 = PC2
So, (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
Now substitute the value of 2x (obtained in equation (1)), weget
7 – 4y + 3y – 6 = 0
– y + 1 = 0
y = 1
By substituting the value of y back in equation (1) we get,
2x = 7 – 4y
2x = 7 – 4(1)
2x = 3
x = 3/2
∴Thepoint P (3/2, 1, 0) in xy-plane is equidistant from A, B and C.
(ii) yz-plane
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
By using the formula,
The distance between any two points (a, b, c) and (m, n, o)is given by,
We know, QA2 = QB2
So, 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
z = [–4y + 7]/4 …. (1)
Since, QA2 = QC2
So, 1 + z2+ (y + 1)2 = (z + 1)2 +(y – 2)2 + 9
2+ 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
Now, substitute the value of z [obtained from (1)] we get
12y + 4y – 7 – 24 = 0
16y – 31 = 0
y = 31/16
Substitute the value of y back in equation (1), we get
∴Thepoint Q (0, 31/16, -3/16) in yz-plane is equidistant from A, B and C.
(iii) zx-plane
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
By using the formula,
The distance between any two points (a, b, c) and (m, n, o)is given by,
We know, RA2 = RB2
So, 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)
Since, RA2 = RC2
So, 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
Substitute the value of 2x [obtained from equation (1)] we get,
2(–4z + 7) = 12 + 2z
–8z + 14 = 12 + 2z
14 – 12 = 8z + 2z
10z = 2
z = 2/10
= 1/5
Now, substitute the value of z back in equation (1), we get
2x = -4z + 7
∴Thepoint R (31/10, 0, 1/5) in xz-plane is equidistant from A, B and C.