To find the equation of a line that is perpendicular to \(y = 37x - 4\) and passes through the point (6, -3), we need to determine the slope of the perpendicular line.
The given equation \(y = 37x - 4\) is in slope-intercept form (\(y = mx + b\)), where \(m\) represents the slope of the line. In this case, the slope of the given line is 37.
Perpendicular lines have slopes that are negative reciprocals of each other. So, the slope of the perpendicular line can be found by taking the negative reciprocal of 37:
\(m_{\perp} = -\frac{1}{37}\)
Now, we have the slope of the perpendicular line. We can use this slope and the given point (6, -3) to determine the equation of the line using the point-slope form of a linear equation:
\(y - y_1 = m_{\perp}(x - x_1)\)
Substituting the values \(x_1 = 6\), \(y_1 = -3\), and \(m_{\perp} = -\frac{1}{37}\), we get:
\(y - (-3) = -\frac{1}{37}(x - 6)\)
Simplifying:
\(y + 3 = -\frac{1}{37}x + \frac{6}{37}\)
Rearranging the equation to the standard form (Ax + By = C):
\(\frac{1}{37}x + y = \frac{6}{37} - 3\)
\(\frac{1}{37}x + y = \frac{6}{37} - \frac{111}{37}\)
\(\frac{1}{37}x + y = -\frac{105}{37}\)
Multiplying both sides of the equation by 37 to eliminate the fraction:
\(x + 37y = -105\)
Therefore, the equation of the line that is perpendicular to \(y = 37x - 4\) and passes through the point (6, -3) is \(x + 37y = -105\).