5. Let \( A(O, 2), B(-1,4), C(-2, x+1) \) and \( D(1,1) \) If \( A C \perp B D \), what is \( x \) ? \( \begin{array}{lllll}\text { A) }-\frac{1}{2} & \text { B) }-\frac{1}{3} & \text { C) }-1 & \text { D) } 1 & \text { E) } \frac{1}{2}\end{array} \)

To solve for \( x \) under the condition that line segment \( AC \) is perpendicular to line segment \( BD \), we need to find the slopes of both lines and set their product equal to -1 (since perpendicular lines have slopes that multiply to -1). 1. The coordinates for points are: - \( A(0, 2) \) - \( C(-2, x+1) \) - \( B(-1, 4) \) - \( D(1, 1) \) 2. **Finding the slope of line \( AC \)**: \[ \text{slope of } AC = \frac{(x + 1) - 2}{-2 - 0} = \frac{x - 1}{-2} = -\frac{x - 1}{2} \] 3. **Finding the slope of line \( BD \)**: \[ \text{slope of } BD = \frac{1 - 4}{1 - (-1)} = \frac{-3}{2} \] 4. **Setting the product of slopes equal to -1**: \[ -\frac{x - 1}{2} \cdot \left(-\frac{3}{2}\right) = -1 \] \[ \frac{(x - 1) \cdot 3}{4} = -1 \] \[ 3(x - 1) = -4 \] \[ 3x - 3 = -4 \] \[ 3x = -1 \] \[ x = -\frac{1}{3} \] Thus, the value of \( x \) is \( \boxed{-\frac{1}{3}} \).