4.3 Q-21

Question Statement

Convert each of the following equations to:

1. Slope-intercept form
2. Intercept form
3. Normal form

Additionally, find the length of the perpendicular from $(0, 0)$ to each line.

Given equations: a. $2x - 4y + 11 = 0$ b. $4x + 7y - 2 = 0$ c. $15y - 8x + 3 = 0$

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Background and Explanation

To solve this, we need to express each equation in three different forms:

1. Slope-intercept form: This is of the form $y = mx + c$, where $m$ is the slope and $c$ is the y-intercept.
2. Intercept form: This is expressed as $\frac{x}{a} + \frac{y}{b} = 1$, where $a$ and $b$ are the x- and y-intercepts, respectively.
3. Normal form: This form is $\frac{Ax}{\sqrt{A^2 + B^2}} + \frac{By}{\sqrt{A^2 + B^2}} = \frac{C}{\sqrt{A^2 + B^2}}$, which expresses the line in terms of its normal vector.

The length of the perpendicular from the origin to the line can be found using the normal form, where the formula is: $\text{Length} = \frac{|C|}{\sqrt{A^2 + B^2}}$

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Solution

a. Equation: $2x - 4y + 11 = 0$

i. Slope-intercept form

Rearrange the equation to solve for $y$:

$$2x - 4y + 11 = 0 \implies 4y = 2x + 11 \implies y = \frac{2}{4}x + \frac{11}{4} \implies y = \frac{1}{4}x + \frac{11}{4}$$

So, the slope-intercept form is: $y = \frac{1}{4}x + \frac{11}{4}$

ii. Intercept form

Rearrange the equation to find the intercepts:

$$2x - 4y = -11 \implies \frac{2x}{-11} + \frac{4y}{-11} = 1 \implies \frac{x}{\frac{11}{2}} + \frac{y}{\frac{11}{4}} = 1$$

So, the intercept form is: $\frac{x}{\frac{11}{2}} + \frac{y}{\frac{11}{4}} = 1$

iii. Normal form

To convert to normal form, first find the length of the normal vector:

$$\text{Normal length} = \sqrt{(2)^2 + (-4)^2} = \sqrt{4 + 16} = \sqrt{20}$$

Then, divide the equation by $\sqrt{20}$:

$$\frac{2}{\sqrt{20}}x - \frac{4}{\sqrt{20}}y = \frac{11}{2\sqrt{5}} \implies \frac{1}{\sqrt{5}}x - \frac{2}{\sqrt{5}}y = \frac{11}{2\sqrt{5}}$$

So, the normal form is: $\frac{1}{\sqrt{5}}x - \frac{2}{\sqrt{5}}y = \frac{11}{2\sqrt{5}}$

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b. Equation: $4x + 7y - 2 = 0$

i. Slope-intercept form

Rearrange the equation to solve for $y$:

$$4x + 7y - 2 = 0 \implies 7y = -4x + 2 \implies y = \frac{-4x + 2}{7} \implies y = \frac{-4}{7}x + \frac{2}{7}$$

So, the slope-intercept form is: $y = \frac{-4}{7}x + \frac{2}{7}$

ii. Intercept form

Rearrange the equation:

$$4x + 7y = 2 \implies \frac{4x}{2} + \frac{7y}{2} = 1 \implies \frac{x}{\frac{1}{2}} + \frac{y}{\frac{2}{7}} = 1$$

So, the intercept form is: $\frac{x}{\frac{1}{2}} + \frac{y}{\frac{2}{7}} = 1$

iii. Normal form

First, calculate the length of the normal vector:

$$\text{Normal length} = \sqrt{(4)^2 + (7)^2} = \sqrt{16 + 49} = \sqrt{65}$$

Then, divide the equation by $\sqrt{65}$:

$$\frac{4}{\sqrt{65}}x + \frac{7}{\sqrt{65}}y = \frac{2}{\sqrt{65}}$$

So, the normal form is: $\frac{4}{\sqrt{65}}x + \frac{7}{\sqrt{65}}y = \frac{2}{\sqrt{65}}$

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c. Equation: $15y - 8x + 3 = 0$

i. Slope-intercept form

Rearrange the equation to solve for $y$:

$$15y - 8x + 3 = 0 \implies 15y = 8x - 3 \implies y = \frac{8}{15}x - \frac{3}{15}$$

So, the slope-intercept form is: $y = \frac{8}{15}x - \frac{3}{15}$

ii. Intercept form

Rearrange the equation:

$$15y - 8x = -3 \implies \frac{-8}{-3}x + \frac{15}{-3}y = \frac{-3}{-3} \implies \frac{x}{3} + \frac{y}{5} = 1$$

So, the intercept form is: $\frac{x}{3} + \frac{y}{5} = 1$

iii. Normal form

First, calculate the length of the normal vector:

$$\text{Normal length} = \sqrt{(8)^2 + (15)^2} = \sqrt{64 + 225} = \sqrt{289} = 17$$

Then, divide the equation by $17$:

$$\frac{-8}{17}x + \frac{15}{17}y = \frac{-3}{17} \implies \frac{8}{17}x - \frac{15}{17}y = \frac{3}{17}$$

So, the normal form is: $\frac{8}{17}x - \frac{15}{17}y = \frac{3}{17}$

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Key Formulas or Methods Used

• Slope-Intercept Form: $y = mx + c$
• Intercept Form: $\frac{x}{a} + \frac{y}{b} = 1$
• Normal Form: $\frac{Ax}{\sqrt{A^2 + B^2}} + \frac{By}{\sqrt{A^2 + B^2}} = \frac{C}{\sqrt{A^2 + B^2}}$
• Length of Perpendicular from the Origin: $\text{Length} = \frac{|C|}{\sqrt{A^2 + B^2}}$

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Summary of Steps

1. Convert the equation to Slope-Intercept Form: Solve for $y$.
2. Convert to Intercept Form: Isolate $x$ and $y$ on opposite sides.
3. Convert to Normal Form: Calculate the normal vector and adjust the equation.
4. Calculate the Length of the Perpendicular using the normal form formula.