The mid-points of the sides of a triangle ABC are given by (-2, 3, 5), (4, -1, 7) and (6, 5, 3). Find the coordinates of A, B and C.

Given: The mid-points of the sides of the triangle are P(-2, 3, 5), Q(4, -1, 7) and R(6, 5, 3).


To find: the coordinates of vertices A, B and C


Formula used:


Section Formula:


A line AB is divided by C in m:n where A(x, y, z) and B(a, b, c).


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The coordinates of C is given by,



We know the mid-point divides side in the ratio of 1:1.


Therefore,


The coordinates of C is given by,




P(-2, 3, 5) is mid-point of A(x1, y1, z1) and B(x2, y2, z2)


Therefore,




Q(4, -1, 7) is mid-point of B(x2, y2, z2) and C(x3, y3, z3)


Therefore,




R(6, 5, 3) is mid-point of A(x1, y1, z1) and C(x3, y3, z3)


Therefore,




x1 + x2 = -4……………………(4)


x2 + x3 = 8………………………(5)


x1 + x3 = 12……………………(6)


Adding (4), (5) and (6):


x1 + x2 + x2 + x3 + x1 + x3 = 8 + 12 – 4


2x1 + 2x2 + 2x3 = 16


2(x1 + x2 + x3) = 16


x1 + x2 + x3 = 8………………………(7)


Subtract (4), (5) and (6) from (7) separately:


x1 + x2 + x3 – x1 – x2 = 8 – (-4)


x3 = 12


x1 + x2 + x3 – x2 – x3 = 8 – 8


x1 = 0


x1 + x2 + x3 – x1 – x3 = 8 – 12


x2 = -4


y1 + y2 = 6……………………(8)


y2 + y3 = -2……………………(9)


y1 + y3 = 10……………………(10)


Adding (8), (9) and (10):


y1 + y2 + y2 + y3 + y1 + y3 = 10 + 6 – 2


2y1 + 2y2 + 2y3 = 14


2(y1 + y2 + y3) = 14


y1 + y2 + y3 = 7………………………(11)


Subtract (8), (9) and (10) from (11) separately:


y1 + y2 + y3 – y1 – y2 = 7 – 6


y3 = 1


y1 + y2 + y3 – y2 – y3 = 7 – (-2)


y1 = 9


y1 + y2 + y3 – y1 – y3 = 7 – 10


y2 = -3


z1 + z2 = 10……………………(12)


z2 + z3 = 14……………………(13)


z1 + z3 = 6……………………(14)


Adding (12), (13) and (14):


z1 + z2 + z2 + z3 + z1 + z3 = 6 + 14 + 10


2z1 + 2z2 + 2z3 = 30


2(z1 + z2 + z3) = 30


z1 + z2 + z3 = 15………………………(15)


Subtract (8), (9) and (10) from (11) separately:


z1 + z2 + z3 – z1 – z2 = 15 – 10


z3 = 5


z1 + z2 + z3 – z2 – z3 = 15 – 14


z1 = 1


z1 + z2 + z3 – z1 – z3 = 15 – 6


z2 = 9


Hence, vertices of sides are A(0, 9, 1) B(-4,-3, 9) and C(12, 1, 5)


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