To find the equation of the line that is perpendicular to [tex]\( y = -4x + 3 \)[/tex] and passes through the point [tex]\((8, 1)\)[/tex], let's follow these steps:
1. Determine the slope of the given line:
The equation [tex]\( y = -4x + 3 \)[/tex] is in the slope-intercept form [tex]\( y = mx + b \)[/tex], where [tex]\( m \)[/tex] is the slope. Here, the slope [tex]\( m = -4 \)[/tex].
2. Find the slope of the perpendicular line:
Lines that are perpendicular to each other have slopes that are negative reciprocals. Therefore, the slope of the line perpendicular to [tex]\( y = -4x + 3 \)[/tex] is the negative reciprocal of [tex]\(-4\)[/tex]. The negative reciprocal of [tex]\(-4 \)[/tex] is [tex]\(\frac{1}{4} \)[/tex].
3. Use the point-slope form to find the equation of the line:
The point-slope form of a line's equation is [tex]\( y - y_1 = m(x - x_1) \)[/tex], where [tex]\( (x_1, y_1) \)[/tex] is a point on the line and [tex]\( m \)[/tex] is the slope. For our problem, we have:
- Point [tex]\((x_1, y_1) = (8, 1)\)[/tex]
- Slope [tex]\( m = \frac{1}{4} \)[/tex]
Plug these values into the point-slope form:
[tex]\[
y - 1 = \frac{1}{4}(x - 8)
\][/tex]
4. Simplify the equation to the slope-intercept form [tex]\( y = mx + b \)[/tex]:
- Distribute the [tex]\(\frac{1}{4}\)[/tex] on the right-hand side:
[tex]\[
y - 1 = \frac{1}{4}x - 2
\][/tex]
- Add 1 to both sides to solve for [tex]\( y \)[/tex]:
[tex]\[
y = \frac{1}{4}x - 1
\][/tex]
So, the equation of the line that is perpendicular to [tex]\( y = -4x + 3 \)[/tex] and goes through the point [tex]\( (8, 1) \)[/tex] is:
[tex]\[
y = \frac{1}{4}x - 1
\][/tex]
In the form [tex]\( y = \frac{a}{b}x + c \)[/tex], the values are:
[tex]\[
a = 1, \quad b = 4, \quad c = -1
\][/tex]
Thus, the desired equation is
[tex]\[
y = \frac{1}{4}x - 1.
\][/tex]
