Q. 4

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).

Answer :

(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


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