Question
Figure ABCD is a parallelogram. If point C lies on the line y = −1, what is the x-value of point C?
coordinate plane with points A negative 2 comma 2, B 2 comma 2, and D 0 comma negative 1, segments AB and AD are drawn, there is also a dashed line from point D on the line y equals negative 1
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coordinate plane with points A negative 2 comma 2, B 2 comma 2, and D 0 comma negative 1, segments AB and AD are drawn, there is also a dashed line from point D on the line y equals negative 1
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2
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Answers
GPT-4o mini
To find the x-value of point C in the parallelogram ABCD, we can use the properties of parallelograms.
Given the points:
- A = (-2, 2)
- B = (2, 2)
- D = (0, -1)
Since ABCD is a parallelogram, the opposite sides must be equal in length and direction, and the diagonals bisect each other.
From points A and B, we have:
- The line segment AB is a horizontal segment from A(-2, 2) to B(2, 2).
Since D is at (0, -1), to find C, we can use the midpoint formula. The midpoint M of diagonal AC should be equal to the midpoint of diagonal BD.
The midpoint M of diagonal AD is:
\[
\text{Midpoint of AD} = \left( \frac{x_A + x_D}{2}, \frac{y_A + y_D}{2} \right) = \left( \frac{-2 + 0}{2}, \frac{2 + (-1)}{2} \right) = \left( -1, \frac{1}{2} \right)
\]
Now, let point C be (x_C, -1). To find the midpoint M of diagonal BC:
\[
\text{Midpoint of BC} = \left( \frac{x_B + x_C}{2}, \frac{y_B + y_C}{2} \right) = \left( \frac{2 + x_C}{2}, \frac{2 + (-1)}{2} \right) = \left( \frac{2 + x_C}{2}, \frac{1}{2} \right)
\]
Setting the midpoints equal:
\[
\left( -1, \frac{1}{2} \right) = \left( \frac{2 + x_C}{2}, \frac{1}{2} \right)
\]
From the second component, we see \( \frac{1}{2} = \frac{1}{2} \), which holds true. Thus, we only need to solve the first component for \( x_C \):
\[
-1 = \frac{2 + x_C}{2}
\]
Multiplying both sides by 2:
\[
-2 = 2 + x_C
\]
Now, isolate \( x_C \):
\[
x_C = -2 - 2 = -4
\]
Thus, point C has coordinates (-4, -1).
The x-value of point C is **-4**.
However, since the problem gives specific options (1, 2, 3, 4) and -4 does not appear among them, please ensure the information you provided is correct, as there might be an error regarding the expected answer or options provided.
Given the points:
- A = (-2, 2)
- B = (2, 2)
- D = (0, -1)
Since ABCD is a parallelogram, the opposite sides must be equal in length and direction, and the diagonals bisect each other.
From points A and B, we have:
- The line segment AB is a horizontal segment from A(-2, 2) to B(2, 2).
Since D is at (0, -1), to find C, we can use the midpoint formula. The midpoint M of diagonal AC should be equal to the midpoint of diagonal BD.
The midpoint M of diagonal AD is:
\[
\text{Midpoint of AD} = \left( \frac{x_A + x_D}{2}, \frac{y_A + y_D}{2} \right) = \left( \frac{-2 + 0}{2}, \frac{2 + (-1)}{2} \right) = \left( -1, \frac{1}{2} \right)
\]
Now, let point C be (x_C, -1). To find the midpoint M of diagonal BC:
\[
\text{Midpoint of BC} = \left( \frac{x_B + x_C}{2}, \frac{y_B + y_C}{2} \right) = \left( \frac{2 + x_C}{2}, \frac{2 + (-1)}{2} \right) = \left( \frac{2 + x_C}{2}, \frac{1}{2} \right)
\]
Setting the midpoints equal:
\[
\left( -1, \frac{1}{2} \right) = \left( \frac{2 + x_C}{2}, \frac{1}{2} \right)
\]
From the second component, we see \( \frac{1}{2} = \frac{1}{2} \), which holds true. Thus, we only need to solve the first component for \( x_C \):
\[
-1 = \frac{2 + x_C}{2}
\]
Multiplying both sides by 2:
\[
-2 = 2 + x_C
\]
Now, isolate \( x_C \):
\[
x_C = -2 - 2 = -4
\]
Thus, point C has coordinates (-4, -1).
The x-value of point C is **-4**.
However, since the problem gives specific options (1, 2, 3, 4) and -4 does not appear among them, please ensure the information you provided is correct, as there might be an error regarding the expected answer or options provided.