Given the equation of the line that we want to find a perpendicular to: [tex]\( y = -\frac{1}{3} x + 1 \)[/tex], we start by identifying the slope of this line.
1. The slope of the given line [tex]\( y = -\frac{1}{3} x + 1 \)[/tex] is [tex]\(-\frac{1}{3}\)[/tex].
2. The slope of a line that is perpendicular to another line is the negative reciprocal of the original line's slope. Therefore, the slope of the perpendicular line is:
[tex]\[ 3 \][/tex]
3. We are given the point [tex]\((6, -1)\)[/tex] through which the perpendicular line should pass.
4. We can use the point-slope form of the equation of a line for this purpose:
[tex]\[ y - y_1 = m (x - x_1) \][/tex]
where [tex]\( m \)[/tex] is the slope and [tex]\((x_1, y_1)\)[/tex] is the point through which the line passes.
Substituting [tex]\( m = 3 \)[/tex] and the point [tex]\((x_1, y_1) = (6, -1) \)[/tex]:
[tex]\[ y - (-1) = 3 (x - 6) \][/tex]
Simplify the equation:
[tex]\[ y + 1 = 3 (x - 6) \][/tex]
So, substituting into the standard point-slope form template provided:
[tex]\[ y - (-1) = 3 (x - 6) \][/tex]
Rewriting it in a familiar point-slope form, we get:
[tex]\[ y - (-1) = 3(x - 6) \][/tex]
Therefore, the final answer is:
[tex]\[ y - (-1) = 3(x - 6) \][/tex]
