Answer :
To determine if a parallelogram [tex]\(ABCD\)[/tex] is a rectangle, we need to examine the properties of its sides and angles. Specifically, in addition to the condition that opposite sides are parallel (which is inherent in any parallelogram), a rectangle requires that adjacent sides are perpendicular to each other. This perpendicularity criterion is mathematically determined by the slopes of the sides.
1. Parallelism of Opposite Sides:
The slopes of opposite sides must be equal for the parallelogram to hold its shape:
[tex]\[ \text{Slope of } AB = \text{Slope of } CD \quad \text{and} \quad \text{Slope of } BC = \text{Slope of } DA. \][/tex]
2. Perpendicularity of Adjacent Sides:
The product of the slopes of two perpendicular lines is [tex]\(-1\)[/tex]. Therefore, the product of the slopes of adjacent sides must be [tex]\(-1\)[/tex].
Let's analyze the options given:
- Option A:
[tex]\[ \left(\frac{y_4-y_3}{x_4-x_3}=\frac{y_3-y_2}{x_3-x_2}\right) \quad \text{and} \quad \left(\frac{y_4-y_3}{x_4-x_3} \times \frac{y_3-y_2}{x_3-x_2}\right)=-1 \][/tex]
This suggests checking the slope of [tex]\(CD\)[/tex] equals the slope of [tex]\(BC\)[/tex] (parallelism), and then verifying they are perpendicular, which doesn't check all adjacent pairs required.
- Option B:
[tex]\[ \left(\frac{y_4-y_3}{x_4-x_3}=\frac{y_2-y_1}{x_2-x_1}\right) \quad \text{and} \quad \left(\frac{y_4-y_3}{x_4-x_3} \times \frac{y_2-y_1}{x_2-x_1}\right)=-1 \][/tex]
This correctly examines if opposite sides (like [tex]\(CD\)[/tex] and [tex]\(AB\)[/tex]) are parallel, and checks if the slopes of [tex]\(CD\)[/tex] and [tex]\(AB\)[/tex] (or any appropriate adjacent sides pair) multiply to [tex]\(-1\)[/tex], ensuring perpendicularity.
- Option C:
[tex]\[ \left(\frac{y_4-y_3}{x_4-x_3}=\frac{y_2-y_1}{x_2-x_1}\right) \quad \text{and} \quad \left(\frac{y_4-y_3}{x_4-x_3} \times \frac{y_3-y_2}{x_3-x_2}\right)=-1 \][/tex]
This closely resembles option B but it incorrectly places perpendicularity check involving non-adjacent slopes.
- Option D:
[tex]\[ \left(\frac{y_4-y_3}{x_4-x_3}=\frac{y_3-y_1}{x_3-x_1}\right) \quad \text{and} \quad \left(\frac{y_4-y_3}{x_4-x_3} \times \frac{y_2-y_1}{x_2-x_1}\right)=-1 \][/tex]
This would imply testing slopes involving different vertices' connections and does not symmetrically validate adjacency/pairing needed for parallelogram sides correctly.
Hence, option B ensures parallelism and perpendicularity in the correct adjacent pairings context verifying [tex]\(ABCD\)[/tex] being a rectangle:
[tex]\[ \left(\frac{y_4-y_3}{x_4-x_3} = \frac{y_2-y_1}{x_2-x_1}\right) \quad \text{and} \quad \left(\frac{y_4-y_3}{x_4-x_3} \times \frac{y_2-y_1}{x_2-x_1}\right) = -1. \][/tex]
Therefore, the correct answer is:
[tex]\[ \boxed{2} \][/tex]
1. Parallelism of Opposite Sides:
The slopes of opposite sides must be equal for the parallelogram to hold its shape:
[tex]\[ \text{Slope of } AB = \text{Slope of } CD \quad \text{and} \quad \text{Slope of } BC = \text{Slope of } DA. \][/tex]
2. Perpendicularity of Adjacent Sides:
The product of the slopes of two perpendicular lines is [tex]\(-1\)[/tex]. Therefore, the product of the slopes of adjacent sides must be [tex]\(-1\)[/tex].
Let's analyze the options given:
- Option A:
[tex]\[ \left(\frac{y_4-y_3}{x_4-x_3}=\frac{y_3-y_2}{x_3-x_2}\right) \quad \text{and} \quad \left(\frac{y_4-y_3}{x_4-x_3} \times \frac{y_3-y_2}{x_3-x_2}\right)=-1 \][/tex]
This suggests checking the slope of [tex]\(CD\)[/tex] equals the slope of [tex]\(BC\)[/tex] (parallelism), and then verifying they are perpendicular, which doesn't check all adjacent pairs required.
- Option B:
[tex]\[ \left(\frac{y_4-y_3}{x_4-x_3}=\frac{y_2-y_1}{x_2-x_1}\right) \quad \text{and} \quad \left(\frac{y_4-y_3}{x_4-x_3} \times \frac{y_2-y_1}{x_2-x_1}\right)=-1 \][/tex]
This correctly examines if opposite sides (like [tex]\(CD\)[/tex] and [tex]\(AB\)[/tex]) are parallel, and checks if the slopes of [tex]\(CD\)[/tex] and [tex]\(AB\)[/tex] (or any appropriate adjacent sides pair) multiply to [tex]\(-1\)[/tex], ensuring perpendicularity.
- Option C:
[tex]\[ \left(\frac{y_4-y_3}{x_4-x_3}=\frac{y_2-y_1}{x_2-x_1}\right) \quad \text{and} \quad \left(\frac{y_4-y_3}{x_4-x_3} \times \frac{y_3-y_2}{x_3-x_2}\right)=-1 \][/tex]
This closely resembles option B but it incorrectly places perpendicularity check involving non-adjacent slopes.
- Option D:
[tex]\[ \left(\frac{y_4-y_3}{x_4-x_3}=\frac{y_3-y_1}{x_3-x_1}\right) \quad \text{and} \quad \left(\frac{y_4-y_3}{x_4-x_3} \times \frac{y_2-y_1}{x_2-x_1}\right)=-1 \][/tex]
This would imply testing slopes involving different vertices' connections and does not symmetrically validate adjacency/pairing needed for parallelogram sides correctly.
Hence, option B ensures parallelism and perpendicularity in the correct adjacent pairings context verifying [tex]\(ABCD\)[/tex] being a rectangle:
[tex]\[ \left(\frac{y_4-y_3}{x_4-x_3} = \frac{y_2-y_1}{x_2-x_1}\right) \quad \text{and} \quad \left(\frac{y_4-y_3}{x_4-x_3} \times \frac{y_2-y_1}{x_2-x_1}\right) = -1. \][/tex]
Therefore, the correct answer is:
[tex]\[ \boxed{2} \][/tex]
