Question: Write a slope-intercept equation for a line passing through the point $(5, -3)$ that is parallel to the line $5x + 7y = 9$. Then write a second equation for a line passing through the point $(5, -3)$ that is perpendicular to the line $5x + 7y = 9$.
The equation of the parallel line is
(Use integers or fractions for any numbers in the equation. Simplify your answer.)
1. **State the problem:** We need to find two lines passing through the point $(5, -3)$: one parallel and one perpendicular to the line $5x + 7y = 9$.
2. **Find the slope of the given line:** Rewrite $5x + 7y = 9$ in slope-intercept form $y = mx + b$.
$$7y = -5x + 9$$
$$y = -\frac{5}{7}x + \frac{9}{7}$$
So, the slope $m$ of the given line is $-\frac{5}{7}$.
3. **Equation of the parallel line:** Parallel lines have the same slope. So the slope of the parallel line is also $-\frac{5}{7}$.
Use point-slope form:
$$y - y_1 = m(x - x_1)$$
where $(x_1, y_1) = (5, -3)$ and $m = -\frac{5}{7}$.
$$y - (-3) = -\frac{5}{7}(x - 5)$$
$$y + 3 = -\frac{5}{7}x + \frac{25}{7}$$
Subtract 3 from both sides:
$$y = -\frac{5}{7}x + \frac{25}{7} - 3$$
Convert 3 to fraction with denominator 7:
$$3 = \frac{21}{7}$$
So,
$$y = -\frac{5}{7}x + \frac{25}{7} - \frac{21}{7} = -\frac{5}{7}x + \frac{4}{7}$$
4. **Equation of the perpendicular line:** The slope of a line perpendicular to one with slope $m$ is the negative reciprocal, so:
$$m_{\perp} = \frac{7}{5}$$
Use point-slope form again with $(5, -3)$:
$$y - (-3) = \frac{7}{5}(x - 5)$$
$$y + 3 = \frac{7}{5}x - 7$$
Subtract 3 from both sides:
$$y = \frac{7}{5}x - 7 - 3$$
$$y = \frac{7}{5}x - 10$$
**Final answers:**
Parallel line: $$y = -\frac{5}{7}x + \frac{4}{7}$$
Perpendicular line: $$y = \frac{7}{5}x - 10$$