# Honors Algebra 2: Midpoints and Perpendicular Lines

Calculator INACTIVE

• Class: Honors Algebra 2
• Author: Peter Atlas
• Text: Algebra and Trigonometry: Structure and Method, Brown

1. Find the midpoint of the segment joining (-2, 4) and (-4, -8).
2. Solution (-3, -2)
3. Find the midpoint of the segment joining $$\displaystyle \left( \frac{3}{7}, -\frac{1}{9} \right)$$ and $$\displaystyle \left( -\frac{5}{2}, -\frac{6}{5} \right)$$.
4. Solution $$\displaystyle \left( -\frac{29}{28}, -\frac{59}{90} \right)$$
5. Find the equation of the line that is the perpendicular bisector of the segment with endpoints (3, -7) and (1, 9).
6. Solution $$\displaystyle y - 1 = \frac{1}{8}(x - 2)$$
7. Find the equation of the line that is the perpendicular bisector of the segment with endpoints (12, 15) and (12, -6).
8. Solution $$\displaystyle y = \frac{9}{2}$$
9. 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 $$\displaystyle \left( -\frac{1}{4}, -\frac{3}{4} \right)$$.
10. Solution $$\displaystyle y + \frac{27}{40} = \frac{15}{11} \left( x + \frac{31}{88} \right)$$
11. Determine an equation of the line passing through (-2, 5) that is perpendicular to the line with equation L: $$3y + 2x = 5$$.
12. Solution $$\displaystyle y - 5 = \frac{3}{2}(x + 2)$$
13. 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.
14. Solution $$33x + 77y = -50$$
15. Find the other endpoint of a segment whose endpoint is $$\displaystyle \left( \frac{1}{4}, -1 \right)$$ and whose midpoint is $$\displaystyle \left( -3, -\frac{5}{8} \right)$$.
16. Solution $$\displaystyle \left( -\frac{25}{4}, -\frac{1}{4} \right)$$
17. 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).
18. 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.