## Lines and Angles Part 2: What Can I Do With Lines and Angles? This lesson is particularly good for seeing relationships and learning how to find missing values based on what you are given. We wrote the Guided Practice with this in mind. We will give you a small bit of information and, from that, you will be able to find multiple other pieces of information.

The axioms of geometry are the building blocks for the entire set of rules, relations, and theorems. Think of the axioms and the inherent properties that naturally come from them as tools in your toolbox. The tools are the proven facts and with these tools you can construct proofs of your own. You can build a self-contained sequence of facts where the first leads directly into the second, that leads directly into the third, and so on until you arrive at your answer.

## Lines and Angles Part 3: Guided Practice The graph above shows 6 lines passing through points A-H with coordinates listed below.

A: $$(\frac{2}{3},\frac{14}{3})$$       E: (1,2)

B: $$(\frac{8}{3},\frac{11}{3})$$       F: (0,1)

C: $$(-\frac{1}{2}, \frac{1}{2})$$       G: (-1,1)

D: (-2,2)       H: (0,0)

1. Which of the following pair line segments are parallel?

A. $$\overline{DA} \parallel \overline{CB},\, and \, \, \overline{DE} \parallel \overline{GF}$$

B. $$\overline{CF} \parallel \overline{EB}, \, and \, \, \overline{GF} \parallel \overline{AB}$$

C. $$\overline{ED} \parallel \overline{AB}, \, and \, \, \overline{HG} \parallel \overline{FG}$$

D. $$\overline{EB} \parallel \overline{GF}, \, and \, \, \overline{FB} \parallel \overline{HC}$$

E. $$\overline{GH} \parallel \overline{CF}, \, and \, \, \overline{GC} \parallel \overline{AD}$$

Answer A has line segment pairs that have the same slope and are therefore parallel.

$$Slope = m = \frac{y_2 – \, y_1}{x_2 – \, x_1}$$

$$\overline{DA} \rightarrow \frac{2 \, -\frac{14}{3}}{-2 \, – \frac{2}{3}} = 1$$

$$\overline{CB} \rightarrow \frac{2 \, – \, \frac{14}{3}}{-2 \, – \, \frac{2}{3}} = 1$$

and

$$\overline{DE} \rightarrow \frac{2 \,- \, 2}{-2 \, – \, 1} = 0$$

$$\overline{GF} \rightarrow \frac{1 \, – \, 1 }{-1 \, – \, 0} = 0$$

1. Which line segments are perpendicular?

A. $$\overline{DA} \perp \overline{CB}, \, and \, \, \overline{DE} \perp \overline{GF}$$

B. $$\overline{CF} \perp \overline{BE}, \, and \, \, \overline{GF} \perp \overline{AB}$$

C. $$\overline{DH} \perp \overline{AB}, \, and \, \, \overline{DE} \perp \overline{GF}$$

D. $$\overline{FE} \perp \overline{FG}, \, and \, \, \overline{FB} \perp \overline{CH}$$

E. $$\overline{GH} \perp \overline{CF},\, and \, \, \overline{GC} \perp \overline{AD}$$

Answer E has line segment pairs that have negative reciprocal (opposite signs and reciprocals) slopes and are therefore perpendicular.

$$\overline{GH} \rightarrow \frac{1 \, – \, 0}{-1 \, – 0} = -1$$

$$\overline{CF} \rightarrow \frac{\frac{1}{2} \, – \, 1}{-\frac{1}{2} \, – \, 0} = 1$$

and

$$\overline{GC} \rightarrow \frac{\frac{1}{2} \,- \, 1}{-\frac{1}{2} \, – \, (-1)} = -1$$

$$\overline{AD} \rightarrow \frac{2 \, -\frac{14}{3}}{-2 \, – \frac{2}{3}} = 1$$

1. How long is line segment $$\overline{DH}$$? $$D = \sqrt{( -2 – 0)^2 + (2 – 0)^2}= \sqrt{4+4} = \sqrt{8}$$

1. What is the midpoint of line segment $$\overline{FE}$$? $$(\frac{0+1}{2},\frac{2+1}{2})= (\frac{1}{2},\frac{3}{2})$$

1. What is the slope of line segment $$\overline{FE}$$?

$$\frac{2-1}{1-0} = 1$$

1. What is the slope of line segment $$\overline{DA}$$?

Because $$\overline{DA}$$ is $$\parallel$$ to $$\overline{FE}$$ its slope is also 1.

1. What is the slope of $$\overline{DH}$$?

Because $$\overline{DH}$$ is $$\perp$$ to $$\overline{DA}$$ and $$\overline{FE}$$ its slope is -1.

1. If $$\angle B$$ is $$60 ^\circ$$, what are the measures of the angles marked A, C, D, E, F, and G?

$$A = 120 ^\circ$$ by the property of same side exterior angles.

$$C = 90 ^\circ$$. $$\overline{DH}$$ and  $$\overline{FE}$$ are $$\perp$$. C is the intersection point of the two line segments. Because perpendicular lines meet at right angles, $$C = 90 ^\circ$$.

$$D = 90 ^\circ$$ by the properties of same-side interior angles and supplementary angles.

Also, $$\overline{DA}$$ is $$\perp$$ to $$\overline{DH}$$ so $$D = 90 ^\circ$$

$$E = 45 ^\circ$$. Since angle C is $$90^\circ$$, the other two angles of $$\triangle CDE$$ must add to $$90^\circ$$. We can use the distance formula to see that the lengths of $$\overline{CE}$$ and $$\overline{CD}$$ are equal, therefore their corresponding angles must be equal as well ($$\triangle CDE$$ is a 45-45-90 triangle). Since $$\angle CED$$ is $$45^\circ$$, $$\angle E = 45 ^\circ$$ by the property of vertical angles.

$$F = 45 ^\circ$$. $$\triangle CDE$$ and $$\triangle CGF$$ are similar triangles (meaning they have proportional sides and equivalent angle measures). Therefore, $$angle CFG = 45^\circ$$ and, by the property of vertical angles, $$\angle F = 45^\circ$$.

$$G = 135 ^\circ$$ We just found that $$\angle CGF = 45^\circ$$, therefore, by the property of supplementary angles, $$\angle G = 135^\circ$$.