1. **State the problem:** Convert the line equation $4x + 7y - 2 = 0$ into three forms: 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: $$4x + 7y - 2 = 0$$ Isolate $y$: $$7y = -4x + 2$$ Divide both sides by 7: $$y = \frac{\cancel{7}7}{\cancel{7}7}y = \frac{-4}{7}x + \frac{2}{7}$$ So, $$y = -\frac{4}{7}x + \frac{2}{7}$$ 3. **Two-intercept form:** This form is $\frac{x}{a} + \frac{y}{b} = 1$, where $a$ and $b$ are the x- and y-intercepts. Find x-intercept by setting $y=0$: $$4x + 7(0) - 2 = 0 \Rightarrow 4x = 2 \Rightarrow x = \frac{2}{4} = \frac{1}{2}$$ Find y-intercept by setting $x=0$: $$4(0) + 7y - 2 = 0 \Rightarrow 7y = 2 \Rightarrow y = \frac{2}{7}$$ Write the two-intercept form: $$\frac{x}{\frac{1}{2}} + \frac{y}{\frac{2}{7}} = 1$$ 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. Calculate $p$: $$p = \frac{|c|}{\sqrt{a^2 + b^2}} = \frac{| -2 |}{\sqrt{4^2 + 7^2}} = \frac{2}{\sqrt{16 + 49}} = \frac{2}{\sqrt{65}}$$ Calculate $\cos \alpha$ and $\sin \alpha$: $$\cos \alpha = \frac{a}{\sqrt{a^2 + b^2}} = \frac{4}{\sqrt{65}}, \quad \sin \alpha = \frac{7}{\sqrt{65}}$$ So the normal form is: $$\frac{4}{\sqrt{65}} x + \frac{7}{\sqrt{65}} y = \frac{2}{\sqrt{65}}$$