Honors Algebra 2: Midpoints and Perpendicular Lines

Calculator INACTIVE

  1. Find the midpoint of the segment joining (-2, 4) and (-4, -8).
    2. Solution (-3, -2)
  2. Find the midpoint of the segment joining \( \displaystyle \left( \frac{3}{7}, -\frac{1}{9} \right) \) and \( \displaystyle \left( -\frac{5}{2}, -\frac{6}{5} \right) \).
    3. Solution \( \displaystyle \left( -\frac{29}{28}, -\frac{59}{90} \right) \)
  3. Find the equation of the line that is the perpendicular bisector of the segment with endpoints (3, -7) and (1, 9).
    4. Solution \( \displaystyle y - 1 = \frac{1}{8}(x - 2) \)
  4. Find the equation of the line that is the perpendicular bisector of the segment with endpoints (12, 15) and (12, -6).
    5. Solution \( \displaystyle y = \frac{9}{2} \)
  5. Find the equation of the line that is the perpendicular bisector of the segment with endpoints \( \displaystyle \left( -\frac{5}{11}, -\frac{3}{5} \right) \) and \( \displaystyle \left( -\frac{1}{4}, -\frac{3}{4} \right) \).
    6. Solution \( \displaystyle y + \frac{27}{40} = \frac{15}{11} \left( x + \frac{31}{88} \right) \)
  6. Determine an equation of the line passing through (-2, 5) that is perpendicular to the line with equation L: \(3y + 2x = 5\).
    7. Solution \( \displaystyle y - 5 = \frac{3}{2}(x + 2) \)
  7. Determine an equation of the line passing through \( \displaystyle \left( -\frac{2}{3}, -\frac{4}{11} \right) \) that is perpendicular to the line L: \(7x - 3y = 10\). Put the answer in standard form.
    8. Solution \( 33x + 77y = -50 \)
  8. Find the other endpoint of a segment whose endpoint is \( \displaystyle \left( \frac{1}{4}, -1 \right) \) and whose midpoint is \( \displaystyle \left( -3, -\frac{5}{8} \right) \).
    9. Solution \( \displaystyle \left( -\frac{25}{4}, -\frac{1}{4} \right) \)
  9. Are the following coordinates vertices of a right triangle? If so, at which vertex is the right angle? Justify your answer. A(1, 2), B(5, 8) and C(2, 10).
    10. Solution Yes. \( \displaystyle m_{AB} = \frac{8 - 2}{5 - 1} = \frac{6}{4} = \frac{2}{3} \). \( \displaystyle m_{BC} = \frac{10 - 8}{2 - 5} = -\frac{2}{3} \). The slopes are negative reciprocals of each other, so the right angle is at B.