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

A. [tex]\(\left(\frac{y_1 - y_1}{x_4 - x_3} = \frac{y_2 - y_2}{x_3 - x_3}\right) \text{ and } \left(\frac{y_4 - y_1}{x_4 - x_3} \times \frac{y_1 - y_2}{z_1 - x_3}\right) = -1\)[/tex]

B. [tex]\(\left(\frac{y_1 - y_2}{x_4 - \alpha_2} = \frac{y_1 - y_1}{x_1 - x_1}\right) \text{ and } \left(\frac{y_1 - y_2}{x_1 - x_1} \times \frac{y_2 - y_2}{y_1 - z_1}\right) = -1\)[/tex]

C. [tex]\(\left(\frac{y_1 - m_1}{x_1 - x_1} = \frac{m_1 - n_1}{z_1 - x_1}\right) \text{ and } \left(\frac{y_1 - y_1}{x_1 - x_1} \times \frac{y_1 - r_1}{z_1 - x_2}\right) = -1\)[/tex]

D. [tex]\(\left(\frac{y_1 - y_2}{x_1 - x_1} = \frac{y_1 - y_2}{x_1 - x_1}\right) \text{ and } \left(\frac{y_1 - y_1}{x_1 - z_1} \times \frac{y_2 - y_1}{x_2 - x_1}\right) = -1\)[/tex]

Sagot :

To establish whether a given parallelogram [tex]\(ABCD\)[/tex] is a rectangle, we have to rely on the geometric property that a rectangle has right angles at each vertex. Mathematically, this can be confirmed if the product of the slopes of the two intersecting sides is [tex]\(-1\)[/tex].

Let's assess each option step by step.

### Option A
[tex]\[ \left(\frac{y_1-y_1}{x_4-x_3}=\frac{y_2-y_2}{x_3-x_3}\right) \, \text{and} \, \left(\frac{y_4-y_1}{x_4-x_3} \times \frac{y_1-y_2}{z_1-x_3}\right)=-1 \][/tex]
- The first condition simplifies incorrect:
[tex]\[\left(\frac{0}{x_4-x_3} = \frac{0}{0}\right)\][/tex]
which is invalid as division by zero is not defined.
- The second condition:
[tex]\(\left(\frac{y_4-y_1}{x_4-x_3} \times \frac{y_1-y_2}{z_1-x_3}\right)=-1\)[/tex]
could be valid in a mathematical context describing perpendicular slopes, but illustrated wrongly.

### Option B
[tex]\[ \left(\frac{y_1-y_2}{x_4-\alpha_2}=\frac{y_1-y_1}{x_1-x_1}\right) \, \text{and} \, \left(\frac{y_1-y_2}{x_1-x_1} \times \frac{y_2-y_2}{y_1-z_1}\right)=-1 \][/tex]
- The first condition incorrectly simplifies to:
[tex]\[\left(\frac{(y_1-y_2)}{(x_4-\alpha_2)} = \frac{0}{0}\right)\][/tex]
which is invalid due to undefined terms.
- The second condition becomes invalid with the product of undefined values (division by zero).

### Option C
[tex]\[ \left(\frac{y_1-m_1}{x_1-x_1}=\frac{m_1-n_1}{z_1-x_1}\right) \, \text{and}\, \left(\frac{y_1-y_1}{x_1-x_1} \times \frac{y_1-r_1}{z_1-x_2}\right)=-1 \][/tex]
- The first condition also simplifies invalid:
[tex]\[\left(\frac{(y_1-m_1)}{0} = \frac{(m_1-n_1)}{(z_1-x_1)}\right)\][/tex]
as division by zero occurs.
- The second product is also invalid due to undefined multiplication of undefined terms.

### Option D
[tex]\[ \left(\frac{y_1-y_2}{x_1-x_1}=\frac{y_1-y_2}{x_1-x_1}\right) \, \text{and}\, \left(\frac{y_1-y_1}{x_1-z_1} \times \frac{y_2-y_1}{x_2-x_1}\right)=-1 \][/tex]
- The first condition invalidly repeats:
[tex]\[\left(\frac{(y_1-y_2)}{0}=\frac{(y_1-y_2)}{0}\right)\][/tex]
as both cases incur division by zero.
- The multiplying slopes:
[tex]\(\left(\frac{0}{x_1-z_1} \times \frac{(y_2-y_1)}{(x_2-x_1)}\right)=-1\)[/tex]
again of mostly undefined basis.

All options incorrectly assert evaluating improper conditions. Correct logic and valid inputs, properly utilize geometric defining, which does not match illustrating steps presented.

Therefore, recognizing implications invalidates all options provided, correct assumption might need verifying steps against typical properties of parallelogram vertices forming rectangle only adhering specific valid approach-wise innocent geometric condition contexts meant re-trial beyond queried steps mismatching bounds specified! Adjust valid question above ensuring plausibly correct options (design ideally relayed required formats identifiable accurately).

Thus, [tex]\( \text{Answer none can ordinary directly legitimate options preserving actual exact properties of rectangle requirements on given information bounds}.\)[/tex]
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