Question

find the equation of the line that is perpendicular to y = (1/8)x - 4 and contains the point (1,-4). y = ?x +

Explanation:

Step1: Find the slope of the perpendicular line

The slope of the given line \( y = \frac{1}{8}x - 4 \) is \( m_1=\frac{1}{8} \). For two perpendicular lines, the product of their slopes is \( -1 \), i.e., \( m_1\times m_2=-1 \). Let the slope of the perpendicular line be \( m_2 \). So, \( \frac{1}{8}\times m_2=-1 \). Solving for \( m_2 \), we get \( m_2 = - 8 \).

Step2: Use point - slope form to find the equation

The point - slope form of a line is \( y - y_1=m(x - x_1) \), where \( (x_1,y_1)=(1,-4) \) and \( m = m_2=-8 \). Substituting these values into the point - slope form, we have \( y-(-4)=-8(x - 1) \).
Simplify the left - hand side: \( y + 4=-8(x - 1) \).
Expand the right - hand side: \( y + 4=-8x+8 \).
Subtract 4 from both sides to get the slope - intercept form (\( y=mx + b \)): \( y=-8x + 8 - 4 \), which simplifies to \( y=-8x+4 \).

Answer:

The equation of the line is \( y=-8x + 4 \), so the coefficient of \( x \) is \(-8\) and the constant term is \( 4 \). Filling in the blanks, we have \( y=\boxed{-8}x+\boxed{4} \).