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Which condition needs to be met to prove [tex][tex]$\overline{A B} \perp \overline{C D}$[/tex][/tex] if [tex][tex]$A\left(x_1, y_1\right)$[/tex][/tex], [tex][tex]$B\left(x_2, y_2\right)$[/tex][/tex], [tex][tex]$C\left(x_3, y_3\right)$[/tex][/tex], and [tex][tex]$D\left(x_4, y_4\right)$[/tex][/tex] form two line segments, [tex][tex]$\overline{A B}$[/tex][/tex] and [tex][tex]$\overline{C D}$[/tex][/tex]?

A. [tex][tex]$\frac{Y_4-Y_2}{X_4-X_2} \times \frac{Y_3-Y_1}{X_3-X_1}=1$[/tex][/tex]

B. [tex][tex]$\frac{Y_4-Y_3}{Y_2-X_1}+\frac{X_4-X_3}{X_2-X_1}=0$[/tex][/tex]

C. [tex][tex]$\frac{Y_4-Y_3}{X_4-X_3} \times \frac{Y_2-Y_1}{X_2-X_1}=-1$[/tex][/tex]

D. [tex][tex]$\frac{Y_2-Y_1}{X_4-X_3}-\frac{X_2-X_1}{Y_4-Y_3}=1$[/tex][/tex]

E. [tex][tex]$\frac{Y_4-Y_2}{Y_2-X_1}+\frac{X_4-X_2}{X_2-X_1}=0$[/tex][/tex]

Sagot :

GinnyAnswer
To determine which condition confirms that two line segments [tex]\(\overline{AB}\)[/tex] and [tex]\(\overline{CD}\)[/tex] are perpendicular to each other, we need to recall a fundamental property of perpendicular lines in a coordinate system.

Two lines are perpendicular if the product of their slopes is [tex]\(-1\)[/tex].

Consider the slopes of the lines [tex]\(\overline{AB}\)[/tex] and [tex]\(\overline{CD}\)[/tex]:

1. The slope of line segment [tex]\(\overline{AB}\)[/tex] can be calculated as:
[tex]\[ m_{AB} = \frac{y_2 - y_1}{x_2 - x_1} \][/tex]

2. The slope of line segment [tex]\(\overline{CD}\)[/tex] can be calculated as:
[tex]\[ m_{CD} = \frac{y_4 - y_3}{x_4 - x_3} \][/tex]

For the line segments [tex]\(\overline{AB}\)[/tex] and [tex]\(\overline{CD}\)[/tex] to be perpendicular, the product of their slopes must equal [tex]\(-1\)[/tex]:
[tex]\[ m_{AB} \times m_{CD} = -1 \][/tex]

Substituting the expressions for the slopes:
[tex]\[ \frac{y_2 - y_1}{x_2 - x_1} \times \frac{y_4 - y_3}{x_4 - x_3} = -1 \][/tex]

Now, we need to identify the correct option from the given choices that matches this condition.

Looking at the options provided:

A. [tex]\(\frac{y_4 - y_2}{x_4 - x_2} \times \frac{y_3 - y_1}{x_3 - x_1} = 1\)[/tex]
- This option does not match our required condition.

B. [tex]\(\frac{y_4 - y_3}{y_2 - x_1} + \frac{x_4 - x_3}{x_2 - x_1} = 0\)[/tex]
- This representation is incorrect and mixes variables in an inconsistent manner.

C. [tex]\(\frac{y_4 - y_3}{x_4 - x_3} \times \frac{y_2 - y_1}{x_2 - x_1} = -1\)[/tex]
- This option matches our derived condition exactly.

D. [tex]\(\frac{y_2 - y_1}{x_4 - x_3} - \frac{x_2 - x_1}{y_4 - y_3} = 1\)[/tex]
- This option does not match the necessary form of having a product of slopes.

E. [tex]\(\frac{y_4 - y_2}{y_2 - x_1} + \frac{x_4 - x_2}{x_2 - x_1} = 0\)[/tex]
- This representation is incorrect and mixes variables in an inconsistent manner.

Thus, the correct condition for proving that the two line segments [tex]\(\overline{AB}\)[/tex] and [tex]\(\overline{CD}\)[/tex] are perpendicular is given by:
[tex]\[ \boxed{\text{C}} \][/tex]
The correct answer is option C.
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