1. **Problem 7:** The value of $y$ varies directly with $x$. Given $x=5$ and $y=-15$, find the constant of variation and the value of $y$ when $x=9$. 2. **Direct Variation Formula:** $y = kx$, where $k$ is the constant of variation. 3. Substitute the known values to find $k$: $$-15 = k \times 5$$ $$k = \frac{-15}{5} = -3$$ 4. So, the constant of variation is $k = -3$. 5. To find $y$ when $x=9$: $$y = -3 \times 9 = -27$$ --- 6. **Problem 9:** Find the equation of the line in slope-intercept form passing through $(9, -3)$ and perpendicular to the line $y = \frac{9}{5}x - 1$. 7. The slope of the given line is $m = \frac{9}{5}$. 8. The slope of a line perpendicular to this is the negative reciprocal: $$m_{\perp} = -\frac{5}{9}$$ 9. Use point-slope form: $$y - y_1 = m_{\perp}(x - x_1)$$ $$y - (-3) = -\frac{5}{9}(x - 9)$$ 10. Simplify: $$y + 3 = -\frac{5}{9}x + 5$$ $$y = -\frac{5}{9}x + 5 - 3$$ $$y = -\frac{5}{9}x + 2$$ 11. The equation is $y = -\frac{5}{9}x + 2$, which corresponds to option B. **Final answers:** - Constant of variation: $-3$ - Value of $y$ when $x=9$: $-27$ - Equation of perpendicular line: $y = -\frac{5}{9}x + 2$ (Option B)