Question

The point (-1, 4) is the image of the point (3, -2) with respect to which one of the following lines?

• A. 2x-3y+1=0
• B. 2x-3y+3=0
• C. 3x-2y-1=0
• D. 3x-2y+1=0

Answer 1

Answer: (A)

2x-3y+1=0

Answer 2

Let P = (3, -2) and P' = (-1, 4). The line of reflection is the perpendicular bisector of the segment PP'. The midpoint M of PP' is: $M = \left( \frac{3 + (-1)}{2}, \frac{-2 + 4}{2} \right) = (1, 1)$ The slope of PP' is: $m_{PP'} = \frac{4 - (-2)}{-1 - 3} = \frac{6}{-4} = -\frac{3}{2}$ The slope of the line of reflection, $m_L$, is the negative reciprocal of $m_{PP'}$: $m_L = -\frac{1}{m_{PP'}} = \frac{2}{3}$ Using the point-slope form $y - y_1 = m(x - x_1)$ with M(1, 1) and $m_L$: $y - 1 = \frac{2}{3}(x - 1)$ $3(y - 1) = 2(x - 1)$ $3y - 3 = 2x - 2$ $2x - 3y + 1 = 0$