Use the Slope and \( y \)-Intercept sliders to graph the line \( y=6 x-2 \). Check the Show 2nd Line box and click and move the red point to graph a parallel line \( L \) through the point \( (2,1) \). Use the interactive figure to find your answer. Use the left and right arrow keys to move along a slider as needed. Click here to launch the interactive figure. Which of the following points are on the line L? A. \( (-4,2) \) B. \( (1,-5) \) OC. \( (-5,1) \) OD. \( (2,3) \)

1. The given line is \( y = 6x - 2 \). A parallel line will have the same slope, so the slope is \( m = 6 \). 2. Line \( L \) is parallel to \( y = 6x - 2 \) and passes through the point \( (2,1) \), so its equation in point-slope form is: \[ y - 1 = 6 (x - 2) \] 3. Convert the above equation to slope-intercept form: \[ y - 1 = 6x - 12 \quad \Longrightarrow \quad y = 6x - 12 + 1 \] \[ y = 6x - 11 \] 4. Now check which of the given points lies on the line \( y = 6x - 11 \). - For point A, \( (-4, 2) \): \[ y = 6(-4) - 11 = -24 - 11 = -35 \quad (\text{Not } 2) \] - For point B, \( (1,-5) \): \[ y = 6(1) - 11 = 6 - 11 = -5 \quad (\text{Matches!}) \] - For point C, \( (-5,1) \): \[ y = 6(-5) - 11 = -30 - 11 = -41 \quad (\text{Not } 1) \] - For point D, \( (2,3) \): \[ y = 6(2) - 11 = 12 - 11 = 1 \quad (\text{Not } 3) \] 5. Only point B, \( (1,-5) \), satisfies the equation \( y = 6x - 11 \). Thus, the point on line \( L \) is \( (1,-5) \).