\begin{tabular}{l} Right triangle \( A B C \) is on a coordinate plane. Segment \( A B \) is on the line \( y=2 \) and is 6 units long. Point \( C \) is on the line \( x=-3 \). If the area of \( \triangle A B C \) is 6 square units, \\ then find a possible \( y \)-coordinate of point \( C \). \\ \begin{tabular}{|}\( O_{5} \) \\ \( O_{5} \) \\ \hline 6\end{tabular} \\ \hline\end{tabular}
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To find a possible \( y \)-coordinate of point \( C \), we start by determining the coordinates of points \( A \) and \( B \). Since segment \( AB \) is 6 units long and lies on the line \( y=2 \), we can place point \( A \) at \((-1, 2)\) and point \( B \) at \( (5, 2) \) (though other configurations are possible, this keeps it simple). Next, we know point \( C \) is on the line \( x = -3 \). Let's denote the coordinates of point \( C \) as \((-3, y_C)\). The area of triangle \( ABC \) can be calculated using the formula: \[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height}. \] In this triangle, the base \( AB \) is 6 units (the distance between points \( A \) and \( B \)) and the height is the vertical distance from point \( C \) to line \( AB \). Since line \( AB \) is at \( y = 2 \), the height will be \( |y_C - 2| \). Given that the area is 6 square units, we can set up the equation: \[ 6 = \frac{1}{2} \times 6 \times |y_C - 2|. \] Simplifying, we find: \[ 6 = 3 \times |y_C - 2| \implies |y_C - 2| = 2. \] This absolute value equation gives us two cases to consider: 1. \( y_C - 2 = 2 \) ⇒ \( y_C = 4 \). 2. \( y_C - 2 = -2 \) ⇒ \( y_C = 0 \). Thus, the possible \( y \)-coordinates of point \( C \) are \( 4 \) and \( 0 \). So, one possible \( y \)-coordinate for point \( C \) is \( \boxed{4} \).
