\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}

To find a possible \( y \)-coordinate of point \( C \) in the right triangle \( ABC \) on the coordinate plane, we can use the formula for the area of a triangle: \[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \] Given that the area of \( \triangle ABC \) is 6 square units, the base is 6 units (since segment \( AB \) is 6 units long), and the height is the distance from point \( C \) to the line \( y = 2 \). Let's denote the \( y \)-coordinate of point \( C \) as \( y_C \). The height of the triangle is the difference between the \( y \)-coordinate of point \( C \) and the line \( y = 2 \), which is \( y_C - 2 \). Substitute the given values into the formula for the area of the triangle: \[ 6 = \frac{1}{2} \times 6 \times (y_C - 2) \] Solve for \( y_C \) to find a possible \( y \)-coordinate of point \( C \). Solve the equation by following steps: - step0: Solve for \(y_{C}\): \(6=\frac{1}{2}\times 6\left(y_{C}-2\right)\) - step1: Multiply the terms: \(6=3\left(y_{C}-2\right)\) - step2: Swap the sides: \(3\left(y_{C}-2\right)=6\) - step3: Divide both sides: \(\frac{3\left(y_{C}-2\right)}{3}=\frac{6}{3}\) - step4: Divide the numbers: \(y_{C}-2=2\) - step5: Move the constant to the right side: \(y_{C}=2+2\) - step6: Add the numbers: \(y_{C}=4\) The possible \( y \)-coordinate of point \( C \) is 4.