Given triangle ABC, segment AB is on the line \( y = 2 \) and has a length of 6 units. Let’s determine the coordinates of points A and B.
Assuming point A is at \( (x_1, 2) \) and point B is at \( (x_2, 2) \) with the distance between them being 6 units: \[ |x_2 - x_1| = 6 \]
Without loss of generality, let’s place point A at \( (0, 2) \) and point B at \( (6, 2) \). Therefore, the coordinates of A and B are: \[ A(0, 2), \quad B(6, 2) \]
Point C is located on the line \( x = -3 \). Let’s denote the coordinates of point C as \( C(-3, y_C) \).
To find the area of triangle ABC, we can use the formula for the area based on the coordinates of the vertices: \[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \] Here, the base \( AB \) is horizontal, and its length is 6. The height is the vertical distance from point C to line \( y = 2 \), which is given by \( |y_C - 2| \).
Thus, the area of triangle ABC can be expressed as: \[ \text{Area} = \frac{1}{2} \times 6 \times |y_C - 2| = 3 |y_C - 2| \]
Given that the area of triangle ABC is 6 square units, we set up the equation: \[ 3 |y_C - 2| = 6 \]
Dividing both sides by 3 gives: \[ |y_C - 2| = 2 \]
This absolute value equation leads to two cases:
- \( y_C - 2 = 2 \) which gives \( y_C = 4 \)
- \( y_C - 2 = -2 \) which gives \( y_C = 0 \)
Now we have found two possible y-coordinates for point C:
- \( y_C = 4 \)
- \( y_C = 0 \)
From the given options \( 3, 4, 5, 6 \), the only possible y-coordinate of point C is: \[ \boxed{4} \]
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