Given the line equation [tex]\( y = 3x - 4 \)[/tex], we need to find the equation of a line that is perpendicular to it and passes through the point [tex]\( (2, 1) \)[/tex].
1. Determine the slope of the given line:
- The given line [tex]\( y = 3x - 4 \)[/tex] is in the slope-intercept form [tex]\( y = mx + b \)[/tex], where [tex]\( m \)[/tex] is the slope.
- The slope [tex]\( m \)[/tex] of the given line is 3.
2. Find the slope of the perpendicular line:
- The slope of a line perpendicular to another line is the negative reciprocal of the original line's slope.
- Therefore, the slope of the perpendicular line is [tex]\( -\frac{1}{3} \)[/tex].
3. Use the point-slope form of the line equation:
- The point-slope form of a line 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.
- We have the point [tex]\( (2, 1) \)[/tex] and the slope [tex]\( -\frac{1}{3} \)[/tex].
- Substituting these into the point-slope form:
[tex]\[
y - 1 = -\frac{1}{3}(x - 2)
\][/tex]
4. Simplify the equation:
- Distribute the slope on the right side:
[tex]\[
y - 1 = -\frac{1}{3} x + \frac{2}{3}
\][/tex]
- Add 1 to both sides to isolate [tex]\( y \)[/tex]:
[tex]\[
y = -\frac{1}{3} x + \frac{2}{3} + 1
\][/tex]
- Combine the constant terms:
[tex]\[
y = -\frac{1}{3} x + \frac{2}{3} + \frac{3}{3}
\][/tex]
[tex]\[
y = -\frac{1}{3} x + \frac{5}{3}
\][/tex]
Therefore, the equation of the line that is perpendicular to the given line and passes through the point [tex]\( (2, 1) \)[/tex] is [tex]\[ y = -\frac{1}{3}x + \frac{5}{3} \][/tex].
The correct answer is:
[tex]\[ \text{B. } y = -\frac{1}{3}x + \frac{5}{3} \][/tex]
