1. **State the problem:** Convert the equation $15y - 8x + 3 = 0$ into slope-intercept form, two-intercept form, and normal form. 2. **Slope-intercept form:** This form is $y = mx + b$, where $m$ is the slope and $b$ is the y-intercept. Start with the given equation: $$15y - 8x + 3 = 0$$ Rearrange to solve for $y$: $$15y = 8x - 3$$ Divide both sides by 15: $$y = \frac{\cancel{15}y}{\cancel{15}} = \frac{8x - 3}{15} = \frac{8}{15}x - \frac{3}{15}$$ Simplify the fraction: $$y = \frac{8}{15}x - \frac{1}{5}$$ 3. **Two-intercept form:** This form is $\frac{x}{a} + \frac{y}{b} = 1$, where $a$ and $b$ are the x- and y-intercepts. Rewrite the original equation: $$15y - 8x + 3 = 0 \implies -8x + 15y = -3$$ Divide both sides by $-3$: $$\frac{-8x}{-3} + \frac{15y}{-3} = \frac{-3}{-3}$$ Simplify: $$\frac{8}{3}x - 5y = 1$$ Rewrite as: $$\frac{x}{\frac{3}{8}} + \frac{y}{-\frac{1}{5}} = 1$$ So the intercepts are: $$a = \frac{3}{8}, \quad b = -\frac{1}{5}$$ 4. **Normal form:** This form is $x \cos \alpha + y \sin \alpha = p$, where $p$ is the distance from the origin to the line and $\alpha$ is the angle between the normal and the x-axis. Start with the original equation: $$15y - 8x + 3 = 0$$ Rewrite as: $$-8x + 15y = -3$$ Calculate the length of the normal vector: $$\sqrt{(-8)^2 + 15^2} = \sqrt{64 + 225} = \sqrt{289} = 17$$ Divide the entire equation by 17: $$\frac{-8}{17}x + \frac{15}{17}y = \frac{-3}{17}$$ Multiply both sides by $-1$ to make $p$ positive: $$\frac{8}{17}x - \frac{15}{17}y = \frac{3}{17}$$ So, $$\cos \alpha = \frac{8}{17}, \quad \sin \alpha = -\frac{15}{17}, \quad p = \frac{3}{17}$$