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: State the problem. Find the equation of the line passing through points $A(0, \frac{3}{4})$ and $B(2\frac{1}{2}, 3\frac{1}{4})$. Step 2: Convert mixed numbers to improper fractions. $2\frac{1}{2} = \frac{5}{2}$ and $3\frac{1}{4} = \frac{13}{4}$. Step 3: Calculate the slope $m$ using the formula: $$m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{\frac{13}{4} - \frac{3}{4}}{\frac{5}{2} - 0} = \frac{\frac{10}{4}}{\frac{5}{2}} = \frac{\frac{5}{2}}{\frac{5}{2}} = 1$$ Step 4: Use point-slope form $y - y_1 = m(x - x_1)$ with point $(0, \frac{3}{4})$: $$y - \frac{3}{4} = 1(x - 0)$$ $$y = x + \frac{3}{4}$$ Step 5: Identify two more points on this line by choosing $x$ values: - For $x=1$, $y=1 + \frac{3}{4} = \frac{7}{4}$ - For $x=3$, $y=3 + \frac{3}{4} = \frac{15}{4}$ Points: $C(1, \frac{7}{4})$, $D(3, \frac{15}{4})$. --- 2. **Write a rule for a line parallel to BC and goes through point (1, \frac{1}{4})**. Step 1: Recall that parallel lines have the same slope. Slope of BC is $1$. 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 $(1, \frac{1}{4})$: $$y - \frac{1}{4} = \frac{1}{4}(x - 1)$$ $$y = \frac{1}{4}x - \frac{1}{4} + \frac{1}{4} = \frac{1}{4}x$$ Step 3: Identify two more points: - For $x=0$, $y=0$ - For $x=4$, $y=1$ Points: $G(0,0)$, $H(4,1)$. --- 4. **Write a rule for a line passing through the origin and lies between BC and GH**. Step 1: BC has slope $1$, GH has slope $\frac{1}{4}$. A line between them has slope $m$ such that $\frac{1}{4} < m < 1$. Step 2: Choose $m=\frac{1}{2}$ for example. Equation: $$y = \frac{1}{2}x$$ --- 5. **Rules and points for given lines:** a. Addition: $x + 1 = y$ Points: $T(0,1)$, $U(1,2)$ b. Line parallel to x-axis: $y = 1 \frac{1}{4} = \frac{5}{4}$ Points: $G(1, \frac{5}{4})$, $H(2, \frac{5}{4})$ c. Multiplication: $5x = y$ Points: $A(1,5)$, $B(2,10)$ d. Line parallel to y-axis: $x = \frac{1}{4}$ Points: $V(\frac{1}{4},0)$, $W(\frac{1}{4},3)$ e. Multiplication with addition: $y = 2x + 1$ Points: $R(1,3)$, $S(2,5)$ --- 6. **Mrs. Boyd's question about lines through (0.6, 1.8):** - Avi's line: $y = 3x$ passes through $(0.6, 1.8)$ because $3 \times 0.6 = 1.8$. - Ezra's line: $x = 0.6$ is vertical and passes through $(0.6, y)$ for any $y$, including $1.8$. - Erik's line: $y = x + 1.2$ passes through $(0.6, 1.8)$ because $0.6 + 1.2 = 1.8$. All these lines contain the point $(0.6, 1.8)$ but have different slopes or orientations. --- **Final answers:** 1. $y = x + \frac{3}{4}$ 2. $y = x - \frac{3}{4}$ 3. $y = \frac{1}{4}x$ 4. $y = \frac{1}{2}x$ 5. See above for each line and points. 6. Explanation above.