Samuel is sketching a design for a t-shirt on some graph paper. He starts with two lines that meet at a right angle. The first line he draws has the equation \( x+2 y=14 \), Select all the possible equations for his second line. A. \( 2 x-y=-7 \) B. \( 7 y=14 x+2 \) C. \( x+2 y=7 \) D. \( 14 y=7 x+2 \) E. \( 14 x+2 y=7 \)

1. Start with the first line: \[ x + 2y = 14. \] Solve for \(y\): \[ 2y = 14 - x \quad \Longrightarrow \quad y = 7 - \frac{1}{2}x. \] The slope of this line is \(-\frac{1}{2}\). 2. For two lines to be perpendicular, the product of their slopes must be \(-1\). Let the slope of the second line be \(m\). Then: \[ \left(-\frac{1}{2}\right) \cdot m = -1 \quad \Longrightarrow \quad m = 2. \] 3. Evaluate each option by rewriting in slope-intercept form \(\big(y = mx + b\big)\): - **Option A:** \(2x - y = -7\) Rearrange to solve for \(y\): \[ -y = -7 - 2x \quad \Longrightarrow \quad y = 7 + 2x. \] Slope \(m = 2\). This line is perpendicular. - **Option B:** \(7y = 14x + 2\) Rearrange: \[ y = \frac{14}{7}x + \frac{2}{7} = 2x + \frac{2}{7}. \] Slope \(m = 2\). This line is perpendicular. - **Option C:** \(x + 2y = 7\) Rearrange: \[ 2y = 7 - x \quad \Longrightarrow \quad y = 7/2 - \frac{1}{2}x. \] Slope \(m = -\frac{1}{2}\). This line is parallel to the first line, not perpendicular. - **Option D:** \(14y = 7x + 2\) Rearrange: \[ y = \frac{7}{14}x + \frac{2}{14} = \frac{1}{2}x + \frac{1}{7}. \] Slope \(m = \frac{1}{2}\). This line is not perpendicular to the first line. - **Option E:** \(14x + 2y = 7\) Rearrange: \[ 2y = 7 - 14x \quad \Longrightarrow \quad y = \frac{7}{2} - 7x. \] Slope \(m = -7\). This line is not perpendicular to the first line. 4. The only options with slope \(2\) (required for perpendicularity) are **Options A and B**.