1. **Problem 1:** Find the standard form of a line through $(-1,-4)$ parallel to the line through $(-2,-5)$ and $(2,7)$. 2. Find the slope of the given line using points $(-2,-5)$ and $(2,7)$: $$m = \frac{7 - (-5)}{2 - (-2)} = \frac{12}{4} = 3$$ 3. Since the new line is parallel, it has the same slope $m=3$. 4. Use point-slope form with point $(-1,-4)$: $$y - (-4) = 3(x - (-1)) \Rightarrow y + 4 = 3(x + 1)$$ 5. Simplify: $$y + 4 = 3x + 3$$ $$y = 3x + 3 - 4$$ $$y = 3x - 1$$ 6. Convert to standard form $Ax + By = C$: $$y = 3x - 1 \Rightarrow -3x + y = -1$$ Multiply both sides by $-1$ to get positive $x$ coefficient: $$\cancel{-3}x + y = -1 \Rightarrow 3x - y = 1$$ 7. **Problem 2:** Find point-slope form of a line through $(2,-8)$ parallel to $y = -\frac{9}{4}x - \frac{1}{2}$. 8. The slope of given line is $m = -\frac{9}{4}$. 9. Use point-slope form: $$y - (-8) = -\frac{9}{4}(x - 2) \Rightarrow y + 8 = -\frac{9}{4}(x - 2)$$ 10. **Problem 3:** Find slope-intercept form of a line parallel to $4x - 10y = -1$ passing through $(15,13)$. 11. Find slope of given line: $$4x - 10y = -1 \Rightarrow -10y = -4x - 1 \Rightarrow y = \frac{4}{10}x + \frac{1}{10} = \frac{2}{5}x + \frac{1}{10}$$ Slope $m = \frac{2}{5}$. 12. Use point-slope form with $(15,13)$: $$y - 13 = \frac{2}{5}(x - 15)$$ 13. Simplify to slope-intercept form: $$y - 13 = \frac{2}{5}x - 6$$ $$y = \frac{2}{5}x - 6 + 13$$ $$y = \frac{2}{5}x + 7$$