How do you find the perpendicular line to y = 3x + 2 through (1, 4)?

To find the perpendicular line to \( y = 3x + 2 \) through \( (1, 4) \), use the negative reciprocal of the slope.

The slope of the given line \( y = 3x + 2 \) is 3. For a line to be perpendicular, its slope must be the negative reciprocal of 3, which is \(-\frac{1}{3}\). This means the slope of the perpendicular line is \(-\frac{1}{3}\).

Next, we use the point-slope form of a line equation, which is \( y - y_1 = m(x - x_1) \), where \( m \) is the slope and \( (x_1, y_1) \) is the point the line passes through. Here, \( m = -\frac{1}{3} \) and the point is \( (1, 4) \).

Substitute these values into the point-slope form:
\[ y - 4 = -\frac{1}{3}(x - 1) \]

Now, simplify the equation:
\[ y - 4 = -\frac{1}{3}x + \frac{1}{3} \]

Add 4 to both sides to get the equation in the slope-intercept form \( y = mx + c \):
\[ y = -\frac{1}{3}x + \frac{1}{3} + 4 \]

Combine the constants:
\[ y = -\frac{1}{3}x + \frac{1}{3} + \frac{12}{3} \]
\[ y = -\frac{1}{3}x + \frac{13}{3} \]

So, the equation of the line perpendicular to \( y = 3x + 2 \) and passing through \( (1, 4) \) is \( y = -\frac{1}{3}x + \frac{13}{3} \).