Let ABCD be a trapezoid with sides AB and CD parallel. let M₁ and M₂ be the midpoints of the nonparallel sides (AD and BC). Use the vector methods taught in class to show that the vector of M₁ and M₂ = 1/2(vector of AB + vector of DC)
M1 = (D-A)/2
M2 = (C-B)/2
the vector from M1 to M2 is
M2-M1 = (C-B)/2 - (D-A)/2 = (C-D)/2 + (A-B)/2
thank you so much
To show that the vector of M₁ and M₂ is equal to half the vector of AB plus DC, we can use the properties of vectors and the midpoint formula.
Let's denote the position vectors of points A, B, C, D, M₁, and M₂ as follows:
- Position vector of A: **
- Position vector of B: **b**
- Position vector of C: **c**
- Position vector of D: **d**
- Position vector of M₁: **m₁**
- Position vector of M₂: **m₂**
First, let's express the vectors AB and DC using their position vectors:
- Vector AB: **AB** = **b** - **
- Vector DC: **DC** = **c** - **d**
Now, let's find the midpoint of the nonparallel sides AD and BC using the midpoint formula:
- Midpoint of AD: **m₁** = (** + **d**) / 2
- Midpoint of BC: **m₂** = (**b** + **c**) / 2
To show that **m₁** + **m₂** = (1/2) (**AB** + **DC**), we need to prove that:
(** + **d**) / 2 + (**b** + **c**) / 2 = (1/2) (**b** - ** + **c** - **d**)
Let's simplify both sides of the equation:
Left-hand side:
- (** + **d**) / 2 + (**b** + **c**) / 2
- (1/2) (** + **b** + **c** + **d**)
Right-hand side:
- (1/2) (**b** - ** + **c** - **d**)
Now, let's distribute the (1/2) factor:
Left-hand side:
- (1/2) ** + (1/2) **b** + (1/2) **c** + (1/2) **d**
Right-hand side:
- (1/2) **b** - (1/2) ** + (1/2) **c** - (1/2) **d**
Now, let's group the terms:
Left-hand side:
- ((1/2) ** - (1/2) **) + ((1/2) **b** + (1/2) **b**) + ((1/2) **c** + (1/2) **c**) + ((1/2) **d** - (1/2) **d**)
- 0 + **b** + **c** + 0
- **b** + **c**
Right-hand side:
- (1/2) **b** - (1/2) ** + (1/2) **c** - (1/2) **d**
As we can see, the left-hand side simplifies to **b** + **c**, which is equal to the right-hand side.
Hence, we have shown that the vector of M₁ and M₂ is equal to (1/2) (**AB** + **DC**), as desired.