Parallel Perpendicular 4Efae7

1. **State the problem:** We need to find two equations of lines passing through the point $(4, -2)$: one parallel and one perpendicular to the line $4x + 7y = 9$. 2. **Find the slope of the given line:** Rewrite $4x + 7y = 9$ in slope-intercept form $y = mx + b$. $$7y = -4x + 9$$ $$y = \frac{-4}{7}x + \frac{9}{7}$$ So, the slope of the given line is $m = -\frac{4}{7}$. 3. **Equation of the parallel line:** Parallel lines have the same slope. So the slope of the parallel line is also $m = -\frac{4}{7}$. Use point-slope form: $$y - y_1 = m(x - x_1)$$ where $(x_1, y_1) = (4, -2)$. $$y - (-2) = -\frac{4}{7}(x - 4)$$ $$y + 2 = -\frac{4}{7}x + \frac{16}{7}$$ Subtract 2 from both sides: $$y = -\frac{4}{7}x + \frac{16}{7} - 2$$ Convert 2 to fraction with denominator 7: $$2 = \frac{14}{7}$$ So, $$y = -\frac{4}{7}x + \frac{16}{7} - \frac{14}{7} = -\frac{4}{7}x + \frac{2}{7}$$ 4. **Equation of the perpendicular line:** The slope of a line perpendicular to slope $m$ is the negative reciprocal: $$m_{\perp} = -\frac{1}{m} = -\frac{1}{-\frac{4}{7}} = \frac{7}{4}$$ Use point-slope form again: $$y - (-2) = \frac{7}{4}(x - 4)$$ $$y + 2 = \frac{7}{4}x - 7$$ Subtract 2 from both sides: $$y = \frac{7}{4}x - 7 - 2$$ $$y = \frac{7}{4}x - 9$$ **Final answers:** - Parallel line: $$y = -\frac{4}{7}x + \frac{2}{7}$$ - Perpendicular line: $$y = \frac{7}{4}x - 9$$