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).
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)