Question

use the given conditions to write an equation for the line in point - slope form and general form. passing through (1, - 4) and perpendicular to the line whose equation is x - 8y - 3 = 0. the equation of the line in point - slope form is \square. (type an equation. use integers or fractions for any numbers in the equation.)

Explanation:

Step1: Find slope of given line

Rewrite $x-8y-3=0$ to slope-intercept form $y=mx+b$:

$$\begin{align*} -8y&=-x+3\\ y&=\frac{1}{8}x-\frac{3}{8} \end{align*}$$

Slope of given line is $m_1=\frac{1}{8}$.

Step2: Find slope of perpendicular line

Perpendicular slopes multiply to $-1$, so:

$$ m_2 = -\frac{1}{m_1} = -8 $$

Step3: Write point-slope form

Use point $(1,-4)$ and $m_2=-8$. Point-slope formula: $y-y_1=m(x-x_1)$

$$ y-(-4) = -8(x-1) $$

Simplify the left side:

$$ y+4 = -8(x-1) $$

Step4: Convert to general form

Expand and rearrange to $Ax+By+C=0$:

$$\begin{align*} y+4&=-8x+8\\ 8x+y+4-8&=0\\ 8x+y-4&=0 \end{align*}$$

Answer:

Point-slope form: $y+4 = -8(x-1)$
General form: $8x+y-4=0$