Objective: I can determine the name of a quadrilateral or a triangle given its coordinate points using slope and distance formulas for the figures: trapezoid, parallelogram, rectangle, rhombus, square, or kite. Prove the best name for each quadrilateral by finding slope and length. Quadrilateral \#1: A $$ (1,2) $$ B $$ (4,5) $$ C $$ (8,1) $$ $$ \mathrm{D}(5,-2) $$ Best Name: $$ \qquad $$ Rectangle Justify on the back $$ \Rightarrow $$. Quadrilateral \#2: $$ \begin{array}{l} \mathrm{E}(-6,2) \ \mathrm{F}(-3,6) \ \mathrm{G}(0,2) \ \mathrm{H}(-3,-5) \end{array} $$ Best Name: $$ \qquad $$ Justify on the back $$ \Rightarrow $$. Quadrilateral \#3: $$ \begin{array}{l} I(-5,2) \ J(1,5) \ K(-6,-6) \ L(8,1) \end{array} $$ Best Name: $$ \qquad $$ Justify on the back $$ \Rightarrow $$. Quadrilateral \#4: $$ \begin{array}{l} \mathrm{M}(0,-3) \ \mathrm{N}(5,0) \ \mathrm{O}(8,-5) \ \mathrm{P}(3,-8) \end{array} $$ Best Name: $$ \qquad $$ Square Justify on the back $$ \Rightarrow $$. Quadrilateral \#5: Q $$ (2,5) $$ $$ R(6,3) $$ $$ S(0,-2) $$ $$ \mathrm{T}(-4,0) $$ Best Name: $$ \qquad $$ Parallelogram Justify on the back $$ \Rightarrow $$.

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- Slope AB: $$ m_{AB} = \frac{5-2}{4-1} = 1 $$
- Slope BC: $$ m_{BC} = \frac{1-5}{8-4} = -1 $$
- Slope CD: $$ m_{CD} = \frac{-2-1}{5-8} = \frac{-3}{-3} = 1 $$
- Slope DA: $$ m_{DA} = \frac{2-(-2)}{1-5} = \frac{4}{-4} = -1 $$

- Length AB: $$ d_{AB} = \sqrt{(4-1)^2 + (5-2)^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2} $$
- Length BC: $$ d_{BC} = \sqrt{(8-4)^2 + (1-5)^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2} $$
- Length CD: $$ d_{CD} = \sqrt{(5-8)^2 + (-2-1)^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2} $$
- Length DA: $$ d_{DA} = \sqrt{(1-5)^2 + (2-(-2))^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2} $$

Since opposite sides are equal in length and the slopes of opposite sides are equal, Quadrilateral #1 is a **Rectangle**.
### Quadrilateral #2: E(-6,2), F(-3,6), G(0,2), H(-3,-5)
- Slope EF: $$ m_{EF} = \frac{6-2}{-3-(-6)} = \frac{4}{3} $$
- Slope FG: $$ m_{FG} = \frac{2-6}{0-(-3)} = \frac{-4}{3} $$
- Slope GH: $$ m_{GH} = \frac{-5-2}{-3-0} = \frac{-7}{-3} = \frac{7}{3} $$
- Slope HE: $$ m_{HE} = \frac{2-(-5)}{-6-(-3)} = \frac{7}{-3} $$

- Length EF: $$ d_{EF} = \sqrt{(-3+6)^2 + (6-2)^2} = \sqrt{9 + 16} = 5 $$
- Length FG: $$ d_{FG} = \sqrt{(0+3)^2 + (2-6)^2} = \sqrt{9 + 16} = 5 $$
- Length GH: $$ d_{GH} = \sqrt{(-3-0)^2 + (-5-2)^2} = \sqrt{9 + 49} = \sqrt{58} $$
- Length HE: $$ d_{HE} = \sqrt{(-6+3)^2 + (2+5)^2} = \sqrt{9 + 49} = \sqrt{58} $$

Since two pairs of opposite sides are equal in length and the slopes of adjacent sides are negative reciprocals, Quadrilateral #2 is a **Kite**.
### Quadrilateral #3: I(-5,2), J(1,5), K(-6,-6), L(8,1)
- Slope IJ: $$ m_{IJ} = \frac{5-2}{1-(-5)} = \frac{3}{6} = \frac{1}{2} $$
- Slope JK: $$ m_{JK} = \frac{-6-5}{-6-1} = \frac{-11}{-7} = \frac{11}{7} $$
- Slope KL: $$ m_{KL} = \frac{1-(-6)}{8-(-6)} = \frac{7}{14} = \frac{1}{2} $$
- Slope LI: $$ m_{LI} = \frac{2-1}{-5-8} = \frac{1}{-13} $$

- Length IJ: $$ d_{IJ} = \sqrt{(1+5)^2 + (5-2)^2} = \sqrt{36 + 9} = 3\sqrt{5} $$
- Length JK: $$ d_{JK} = \sqrt{(-6-1)^2 + (-6-5)^2} = \sqrt{49 + 121} = \sqrt{170} $$
- Length KL: $$ d_{KL} = \sqrt{(8+6)^2 + (1-(-6))^2} = \sqrt{196 + 49} = 7\sqrt{7} $$
- Length LI: $$ d_{LI} = \sqrt{(-5-8)^2 + (2-1)^2} = \sqrt{169 + 1} = \sqrt{170} $$

Since two pairs of opposite sides are equal in length and the slopes of opposite sides are equal, Quadrilateral #3 is a **Parallelogram**.
### Quadrilateral #4: M(0,-3), N(5,0), O(8,-5), P(3,-8)
- Slope MN: $$ m_{MN} = \frac{0-(-3)}{5-0} = \frac{3}{5} $$
- Slope NO: $$ m_{NO} = \frac{-5-0}{8-5} = \frac{-5}{3} $$
- Slope OP: $$ m_{OP} = \frac{-8-(-5)}{3-8} = \frac{-3}{-5} = \frac{3}{5} $$
- Slope PM: $$ m_{PM} = \frac{-3-(-8)}{0-3} = \frac{5}{-3} $$

- Length MN: $$ d_{MN} = \sqrt{(5-0)^2 + (0-(-3))^2} = \sqrt{25 + 9} = \sqrt{34} $$
- Length NO: $$ d_{NO} = \sqrt{(8-5)^2 + (-5-0)^2} = \sqrt{9 + 25} = \sqrt{34} $$
- Length OP: $$ d_{OP} = \sqrt{(3-8)^2 + (-8-(-5))^2} = \sqrt{25 + 9} = \sqrt{34} $$
- Length PM: $$ d_{PM} = \sqrt{(0-3)^2 + (-3-(-8))^2} = \sqrt{9 + 25} = \sqrt{34} $$

Since all sides are equal in length and the slopes of opposite sides are equal, Quadrilateral #4 is a **Square**.
### Quadrilateral #5: Q(2,5), R(6,3), S(0,-2), T(-4,0)
**Quadrilateral #1:** Rectangle **Quadrilateral #2:** Kite **Quadrilateral #3:** Parallelogram **Quadrilateral #4:** Square **Quadrilateral #5:** Parallelogram

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