1. **Problem Statement:** Transform the equation $-2x + 5y = 10$ into four different forms: (i) Two points form, (ii) Two intercepts form, (iii) Symmetric form, and (iv) Normal form. 2. **Given Equation:** $$-2x + 5y = 10$$ 3. **(i) Two Points Form:** Choose two points on the line by assigning values to $x$ and solving for $y$. - Let $x=0$: $$-2(0) + 5y = 10 \Rightarrow 5y = 10 \Rightarrow y = 2$$ - Let $y=0$: $$-2x + 5(0) = 10 \Rightarrow -2x = 10 \Rightarrow x = -5$$ Two points are $(0,2)$ and $(-5,0)$. The two points form is: $$y - 2 = \frac{0 - 2}{-5 - 0}(x - 0) = \frac{-2}{-5}x = \frac{2}{5}x$$ Simplified: $$y - 2 = \frac{2}{5}x$$ 4. **(ii) Two Intercepts Form:** The intercepts are $x$-intercept $(-5,0)$ and $y$-intercept $(0,2)$. The two intercepts form is: $$\frac{x}{-5} + \frac{y}{2} = 1$$ 5. **(iii) Symmetric Form:** Rewrite the equation to isolate $x$ and $y$ terms: $$-2x + 5y = 10 \Rightarrow 5y = 10 + 2x \Rightarrow y = 2 + \frac{2}{5}x$$ From intercepts: $$\frac{x}{-5} + \frac{y}{2} = 1 \Rightarrow \frac{x}{-5} = 1 - \frac{y}{2}$$ Symmetric form is: $$\frac{x}{-5} + \frac{y}{2} = 1$$ 6. **(iv) Normal Form:** The normal form is: $$x \cos \alpha + y \sin \alpha = p$$ Rewrite original equation: $$-2x + 5y = 10$$ Calculate length of normal vector: $$\sqrt{(-2)^2 + 5^2} = \sqrt{4 + 25} = \sqrt{29}$$ Divide entire equation by $\sqrt{29}$: $$\frac{-2}{\sqrt{29}}x + \frac{5}{\sqrt{29}}y = \frac{10}{\sqrt{29}}$$ So, $$\cos \alpha = \frac{-2}{\sqrt{29}}, \quad \sin \alpha = \frac{5}{\sqrt{29}}, \quad p = \frac{10}{\sqrt{29}}$$ **Final answers:** (i) Two points form: $$y - 2 = \frac{2}{5}x$$ (ii) Two intercepts form: $$\frac{x}{-5} + \frac{y}{2} = 1$$ (iii) Symmetric form: $$\frac{x}{-5} + \frac{y}{2} = 1$$ (iv) Normal form: $$\frac{-2}{\sqrt{29}}x + \frac{5}{\sqrt{29}}y = \frac{10}{\sqrt{29}}$$