(i) (0, 7, –10), (1, 6, –6), and (4, 9, –6) are the vertices of an isosceles triangle.
Let us consider the points
P(0, 7, –10), Q(1, 6, – 6) and R(4, 9, – 6)
If any 2 sides are equal, it will be an isosceles triangle
So, first, let us calculate the distance of PQ, QR
Calculating PQ
P ≡ (0, 7, – 10) and Q ≡ (1, 6, – 6)
By using the formula,
Distance PQ = √[(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2]
So, here
x1 = 0, y1 = 7, z1 = –10
x2 = 1, y2 = 6, z2 = –6
Distance PQ = √[(1 – 0)2 + (6 – 7)2 + (-6 – (-10))2]
= √[(1)2 + (-1)2 + (4)2]
= √[1 + 1 + 16]
= √18
Calculating QR
Q ≡ (1, 6, –6) and R ≡ (4, 9, –6)
By using the formula,
Distance QR = √[(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2]
So, here
x1 = 1, y1 = 6, z1 = – 6
x2 = 4, y2 = 9, z2 = – 6
Distance QR = √[(4 – 1)2 + (9 – 6)2 + (-6 – (-6))2]
= √[(3)2 + (3)2 + (-6+6)2]
= √[9 + 9 + 0]
= √18
Hence, PQ = QR
18 = 18
2 sides are equal
∴ PQR is an isosceles triangle.
(ii) (0, 7, 10), (–1, 6, 6), and (–4, 9, 6) are the vertices of a right-angled triangle.
Let the points be
P(0, 7, 10), Q(– 1, 6, 6) & R(– 4, 9, 6)
First, let us calculate the distance of PQ, OR and PR
Calculating PQ
P ≡ (0, 7, 10), and Q ≡ (– 1, 6, 6)
By using the formula,
Distance PQ = √[(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2]
So, here
x1 = 0, y1 = 7, z1 = 10
x2 = – 1, y2 = 6, z2 = 6
Distance PQ = √[(-1 – 0)2 + (6 – 7)2 + (6 – 10)2]
= √[(-1)2 + (-1)2 + (-4)2]
= √[1 + 1 + 16]
= √18
Calculating QR
Q ≡ (1, 6, – 6), and R ≡ (4, 9, –6)
By using the formula,
Distance QR = √[(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2]
So, here
x1 = 1, y1 = 6, z1 = – 6
x2 = 4, y2 = 9, z2 = – 6
Distance QR = √[(4 – 1)2 + (9 – 6)2 + (-6 – (-6))2]
= √[(3)2 + (3)2 + (-6+6)2]
= √[9 + 9 + 0]
= √18
Calculating PR
P ≡ (0, 7, 10) and R ≡ (–4, 9, 6)
By using the formula,
Distance PR = √[(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2]
So, here
x1 = 0, y1 = 7, z1 = 10
x2 = – 4, y2 = 9, z2 = 6
Distance PR = √[(-4 – 0)2 + (9 – 7)2 + (6 – 10)2]
= √[(-4)2 + (2)2 + (-4)2]
= √[16 + 4 + 16]
= √36
Now,
PQ2 + QR2 = 18 + 18
= 36
= PR2
By using the converse of Pythagoras theorem,
The given vertices P, Q & R are the vertices of a right-angled triangle at Q.
(iii) (–1, 2, 1), (1, –2, 5), (4, –7, 8), and (2, –3, 4) are the vertices of a parallelogram.
Let the points be A(–1, 2, 1), B(1, –2, 5), C(4, –7, 8), and D(2, –3, 4)
ABCD can be vertices of parallelogram only if opposite sides are equal.
i.e., AB = CD, and BC = AD
First, let us calculate the distance
Calculating AB
A ≡ (– 1, 2, 1) and B ≡ (1, – 2, 5)
By using the formula,
Distance AB = √[(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2]
So, here
x1 = –1, y1 = 2, z1 = 1
x2 = 1, y2 = –2, z2 = 5
Distance AB = √[(1 – (-1))2 + (-2 – 2)2 + (5 – 1)2]
= √[(2)2 + (-4)2 + (4)2]
= √[4 + 16 + 16]
= √36
= 6
Calculating BC
B ≡ (1, –2, 5) and C ≡ (4, –7, 8)
By using the formula,
Distance BC = √[(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2]
So, here
x1 = 1, y1 = –2, z1 = 5
x2 = 4, y2 = –7, z2 = 8
Distance BC = √[(4 – 1)2 + (-7 – (-2))2 + (8 – 5)2]
= √[(3)2 + (-5)2 + (3)2]
= √[9 + 25 + 9]
= √43
Calculating CD
C ≡ (4, –7, 8) and D ≡ (2, –3, 4)
By using the formula,
Distance CD = √[(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2]
So, here
x1 = 4, y1 = – 7, z1 = 8
x2 = 2, y2 = – 3, z2 = 4
Distance CD = √[(2 – 4)2 + (-3 – (-7))2 + (4 – 8)2]
= √[(-2)2 + (4)2 + (-4)2]
= √[4 + 16 + 16]
= √36
= 6
Calculating DA
D ≡ (2, –3, 4) and A ≡ (–1, 2, 1)
By using the formula,
Distance DA = √[(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2]
So, here
x1 = 2, y1 = – 3, z1 = 4
x2 = – 1, y2 = 2, z2 = 1
Distance DA = √[(-1 – 2)2 + (2 – (-3))2 + (1 – 4)2]
= √[(-3)2 + (5)2 + (-3)2]
= √[9 + 25 + 9]
= √43
Since AB = CD and BC = DA (given),
In ABCD, both pairs of opposite sides are equal.
∴ ABCD is a parallelogram.