Verify the following: (–1, 2, 1), (1, –2, 5), (4, –7, 8) and (2, –3, 4) are the vertices of a parallelogram.

Asked by Abhisek | 1 year ago |  99

##### Solution :-

(–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) & D(2, –3, 4)

ABCD can be vertices of parallelogram only if opposite sides are equal.

AB = CD and BC = AD

Firstly let us calculate the distance

Calculating AB

A ≡ (– 1, 2, 1) and B ≡ (1, – 2, 5)

By using the formula,

Distance AB =

$$\sqrt{(x^2 – x^1)^2 + (y^2 – y^1)^2 + (z^2 – z^1)^2}$$

So here,

x1 = – 1, y1 = 2, z1 = 1

x2 = 1, y2 = – 2, z2 = 5

Distance AB =

$$\sqrt{(1 – (-1))^2 + (-2 – 2)^2 + (5 – 1)^2}$$

$$\sqrt{(2)^2 + (-4)^2 + (4)^2}$$

$$\sqrt{4 + 16 + 16}$$

$$\sqrt{36}$$

= 6

Calculating BC

B ≡ (1, – 2, 5) and C ≡ (4, – 7, 8)

By using the formula,

Distance BC =

$$\sqrt{(x^2 – x^1)^2 + (y^2 – y^1)^2 + (z^2 – z^1)^2}$$

So here,

x1 = 1, y1 = – 2, z1 = 5

x2 = 4, y2 = – 7, z2 = 8

Distance BC =

$$\sqrt{(4 – 1)^2 + (-7 – (-2))^2 + (8 – 5)^2}$$

$$\sqrt{(3)^2 + (-5)^2 + (3)^2}$$

$$\sqrt{9 + 25 + 9}$$

$$\sqrt{43}$$

Calculating CD

C ≡ (4, – 7, 8) and D ≡ (2, – 3, 4)

By using the formula,

Distance CD =

$$\sqrt{(x^2 – x^1)^2 + (y^2 – y^1)^2 + (z^2 – z^1)^2}$$

So here,

x1 = 4, y1 = – 7, z1 = 8

x2 = 2, y2 = – 3, z2 = 4

Distance CD =

$$\sqrt{(2 – 4)^2 + (-3 – (-7))^2 + (4 – 8)^2}$$

$$\sqrt{(-2)^2 + (4)^2 + (-4)^2}$$

$$\sqrt{4 + 16 + 16}$$

$$\sqrt{36}$$

= 6

Calculating DA

D ≡ (2, – 3, 4) and A ≡ (– 1, 2, 1)

By using the formula,

Distance DA =

$$\sqrt{(x^2 – x^1)^2 + (y^2 – y^1)^2 + (z^2 – z^1)^2}$$

So here,

x1 = 2, y1 = – 3, z1 = 4

x2 = – 1, y2 = 2, z2 = 1

Distance DA =

$$\sqrt{(-1 – 2)^2 + (2 – (-3))^2 + (1 – 4)^2}$$

$$\sqrt{(-3)^2 + (5)^2 + (-3)^2}$$

$$\sqrt{9 + 25 + 9}$$

$$\sqrt{43}$$

Since AB = CD and BC = DA (given)

So, In ABCD both pairs of opposite sides are equal.

ABCD is a parallelogram.

Answered by Pragya Singh | 1 year ago

### Related Questions

#### A(1, 2, 3), B(0, 4, 1), C(-1, -1, -3) are the vertices of a triangle ABC. Find the point in which the bisector of the

A(1, 2, 3), B(0, 4, 1), C(-1, -1, -3) are the vertices of a triangle ABC. Find the point in which the bisector of the angle ∠BAC meets BC.

#### 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

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.

#### If the points A(3, 2, -4), B(9, 8, -10) and C(5, 4, -6) are collinear, find the ratio in which C divided AB.

If the points A(3, 2, -4), B(9, 8, -10) and C(5, 4, -6) are collinear, find the ratio in which C divided AB.