1. **Write a rule for the line that contains the points (0, \frac{3}{4}) and (2 \frac{1}{2}, 3 \frac{1}{4})** Step 1: Identify the points as $ (0, \frac{3}{4}) $ and $ (2.5, 3.25) $. Step 2: Find the slope $ m $ using the formula: $$ m = \frac{y_2 - y_1}{x_2 - x_1} $$ $$ m = \frac{3.25 - 0.75}{2.5 - 0} = \frac{2.5}{2.5} = 1 $$ Step 3: Use point-slope form to write the equation: $$ y - y_1 = m(x - x_1) $$ Using point $ (0, \frac{3}{4}) $: $$ y - \frac{3}{4} = 1(x - 0) $$ $$ y = x + \frac{3}{4} $$ Step 4: Identify 2 more points on this line by choosing values for $ x $: - For $ x = 1 $, $ y = 1 + \frac{3}{4} = \frac{7}{4} = 1.75 $ - For $ x = 3 $, $ y = 3 + \frac{3}{4} = \frac{15}{4} = 3.75 $ Points B and C: - B: $ (1, \frac{7}{4}) $ - C: $ (3, \frac{15}{4}) $ 2. **Write a rule for a line parallel to BC and goes through point (1, \frac{1}{4})** Step 1: Since BC has slope 1 (from above), a parallel line has the same slope $ m = 1 $. Step 2: Use point-slope form with point $ (1, \frac{1}{4}) $: $$ y - \frac{1}{4} = 1(x - 1) $$ $$ y = x - 1 + \frac{1}{4} = x - \frac{3}{4} $$ 3. **Create a rule for the line that contains points (1, \frac{1}{4}) and (3, \frac{3}{4})** Step 1: Calculate slope: $$ m = \frac{\frac{3}{4} - \frac{1}{4}}{3 - 1} = \frac{\frac{2}{4}}{2} = \frac{\frac{1}{2}}{2} = \frac{1}{4} $$ Step 2: Use point-slope form with point $ (1, \frac{1}{4}) $: $$ y - \frac{1}{4} = \frac{1}{4}(x - 1) $$ $$ y = \frac{1}{4}x - \frac{1}{4} + \frac{1}{4} = \frac{1}{4}x $$ Step 3: Identify 2 more points on this line: - For $ x = 0 $, $ y = 0 $ - For $ x = 4 $, $ y = 1 $ Points G and H: - G: $ (0, 0) $ - H: $ (4, 1) $ 4. **Write a rule for a line that passes through the origin and lies between BC and GH** Step 1: BC has slope 1, GH has slope $ \frac{1}{4} $. Step 2: A line between them has slope between $ \frac{1}{4} $ and 1, for example $ m = \frac{1}{2} $. Step 3: Since it passes through origin: $$ y = \frac{1}{2}x $$ 5. **Create rules for lines containing point $ (\frac{1}{4}, 1 \frac{1}{4}) $ with given operations:** - a. Addition: $ x + 1 = y $ - b. Parallel to x-axis: $ y = 1 \frac{1}{4} = \frac{5}{4} $ - c. Multiplication: $ 4x = y $ - d. Parallel to y-axis: $ x = \frac{1}{4} $ - e. Multiplication with addition: $ 4x + 1 = y $ 6. **Mrs. Boyd's question about lines through (0.6, 1.8):** - Avi's rule: multiply x by 3, so $ y = 3x $ passes through (0.6, 1.8) because $ 3 \times 0.6 = 1.8 $. - Ezra's rule: vertical line $ x = 0.6 $ passes through (0.6, 1.8). - Erik's rule: add 1.2 to x, so $ y = x + 1.2 $ passes through (0.6, 1.8) because $ 0.6 + 1.2 = 1.8 $. All these lines contain the point (0.6, 1.8) because the point satisfies each equation. --- **Final answers:** 1. $ y = x + \frac{3}{4} $ 2. $ y = x - \frac{3}{4} $ 3. $ y = \frac{1}{4}x $ 4. $ y = \frac{1}{2}x $ 5a. $ y = x + 1 $ 5b. $ y = \frac{5}{4} $ 5c. $ y = 4x $ 5d. $ x = \frac{1}{4} $ 5e. $ y = 4x + 1 $ 6. Lines: $ y = 3x $, $ x = 0.6 $, $ y = x + 1.2 $