7. through \( (-1,4) \), perpendicular to \( y=-\frac{1}{8} x+3 \)

To find the equation of a line that passes through the point \((-1, 4)\) and is perpendicular to the line given by the equation \(y = -\frac{1}{8}x + 3\), we first need to determine the slope of the given line. The slope of the line \(y = -\frac{1}{8}x + 3\) is \(-\frac{1}{8}\). For two lines to be perpendicular, the product of their slopes must equal \(-1\). Therefore, if the slope of the given line is \(m_1 = -\frac{1}{8}\), the slope of the line we want to find, \(m_2\), can be calculated as follows: \[ m_1 \cdot m_2 = -1 \] Substituting \(m_1\): \[ -\frac{1}{8} \cdot m_2 = -1 \] To solve for \(m_2\), we multiply both sides by \(-1\): \[ \frac{1}{8} \cdot m_2 = 1 \] Now, multiply both sides by \(8\): \[ m_2 = 8 \] Now that we have the slope of the perpendicular line, which is \(8\), we can use the point-slope form of the equation of a line, which is given by: \[ y - y_1 = m(x - x_1) \] Here, \((x_1, y_1) = (-1, 4)\) and \(m = 8\). Substituting these values into the equation gives: \[ y - 4 = 8(x + 1) \] Now, we can simplify this equation: \[ y - 4 = 8x + 8 \] Adding \(4\) to both sides: \[ y = 8x + 12 \] Thus, the equation of the line that passes through the point \((-1, 4)\) and is perpendicular to the line \(y = -\frac{1}{8}x + 3\) is: \[ \boxed{y = 8x + 12} \]