Select the correct answer.

The vertices of a parallelogram are [tex]\(A\left(x_1, y_1\right), B\left(x_2, y_2\right), C\left(x_3, y_3\right),\)[/tex] and [tex]\(D\left(x_4, y_4\right)\)[/tex]. Which of the following must be true if parallelogram [tex]\(ABCD\)[/tex] is proven to be a rectangle?

A. [tex]\(\left(\frac{y_4-y_1}{z_4-z_3}=\frac{y_1-y_2}{z_3-z_2}\right)\)[/tex] and [tex]\(\left(\frac{y_4-y_2}{z_4-x_3} \times \frac{y_1-y_2}{x_3-x_2}\right)=-1\)[/tex]

B. [tex]\(\left(\frac{y_4-y_3}{x_4-z_3}=\frac{y_2-y_1}{x_2-z_1}\right)\)[/tex] and [tex]\(\left(\frac{y_4-y_3}{x_4-x_3} \times \frac{y_2-y_1}{x_2-x_1}\right)=-1\)[/tex]

C. [tex]\(\left(\frac{y_4-y_3}{x_4-x_3}=\frac{y_2-y_1}{x_2-x_1}\right)\)[/tex] and [tex]\(\left(\frac{y_4-y_3}{x_4-x_3} \times \frac{y_3-y_2}{x_3-x_2}\right)=-1\)[/tex]

D. [tex]\(\left(\frac{y_1-y_3}{x_4-x_3}=\frac{y_3-y_1}{x_3-x_1}\right)\)[/tex] and [tex]\(\left(\frac{y_4-y_3}{z_4-z_3} \times \frac{y_1-y_1}{x_2-x_1}\right)=-1\)[/tex]



Answer :

To determine which condition must be true for a parallelogram [tex]\(ABCD\)[/tex] to be a rectangle, we need to use the properties of rectangles. Specifically, a rectangle has perpendicular sides. In coordinate geometry, two lines are perpendicular if and only if the product of their slopes is [tex]\(-1\)[/tex].

Given the vertices of the parallelogram [tex]\(A(x_1, y_1)\)[/tex], [tex]\(B(x_2, y_2)\)[/tex], [tex]\(C(x_3, y_3)\)[/tex], and [tex]\(D(x_4, y_4)\)[/tex], let's analyze the given choices one by one:

1. Option A:
[tex]\[ \left(\frac{y_4 - y_1}{z_4 - z_3} = \frac{y_1 - y_2}{z_3 - z_2}\right) \quad \text{and} \quad \left(\frac{y_4 - y_2}{z_4 - x_3} \times \frac{y_1 - y_2}{x_3 - x_2}\right) = -1 \][/tex]
This option is incorrect because the conditions and notation do not match the properties of perpendicular slopes.

2. Option B:
[tex]\[ \left(\frac{y_4 - y_3}{x_4 - z_3} = \frac{y_2 - y_1}{x_2 - z_1}\right) \quad \text{and} \quad \left(\frac{y_4 - y_3}{x_4 - x_3} \times \frac{y_2 - y_1}{x_2 - x_1}\right) = -1 \][/tex]
This option has mismatched coordinates and unnecessary complexity that does not correctly address perpendicularity.

3. Option C:
[tex]\[ \left(\frac{y_4 - y_3}{x_4 - x_3} = \frac{y_2 - y_1}{x_2 - x_1}\right) \quad \text{and} \quad \left(\frac{y_4 - y_3}{x_4 - x_3} \times \frac{y_2 - y_1}{x_2 - x_1}\right) = -1 \][/tex]
This option correctly states the condition for perpendicularity:
- First, the slopes of two opposite sides are equal: [tex]\(\frac{y_4 - y_3}{x_4 - x_3} = \frac{y_2 - y_1}{x_2 - x_1}\)[/tex].
- Second, the product of the slopes must be [tex]\(-1\)[/tex] for perpendicularity: [tex]\(\left(\frac{y_4 - y_3}{x_4 - x_3} \times \frac{y_2 - y_1}{x_2 - x_1}\right) = -1\)[/tex].

4. Option D:
[tex]\[ \left(\frac{y_1 - y_3}{x_4 - x_3} = \frac{y_3 - y_1}{x_3 - x_1}\right) \quad \text{and} \quad \left(\frac{y_4 - y_3}{z_4 - z_3} \times \frac{y_1 - y_1}{x_2 - x_1}\right) = -1 \][/tex]
This option is incorrect due to incorrect and inconsistent coordinate expressions and conditions.

To summarize, the correct answer is:

Option C:
[tex]\[ \left(\frac{y_4 - y_3}{x_4 - x_3} = \frac{y_2 - y_1}{x_2 - x_1}\right) \quad \text{and} \quad \left(\frac{y_4 - y_3}{x_4 - x_3} \times \frac{y_2 - y_1}{x_2 - x_1}\right) = -1 \][/tex]

So, the correct choice is:

[tex]\[ \boxed{C} \][/tex]