line p has an equation of y = -8x + 6. line q, which is perpendicular to line p, includes the point (2, -2)…

Explanation:

Step1: Find the slope of line ( q )

The equation of line ( p ) is ( y=-8x + 6 ), and its slope ( m_p=-8 ). For two perpendicular lines with slopes ( m_1 ) and ( m_2 ), ( m_1\times m_2=-1 ). Let the slope of line ( q ) be ( m_q ). So, ( -8\times m_q=-1 ), then ( m_q=\frac{1}{8} ).

Step2: Use the point - slope form to find the equation of line ( q )

The point - slope form of a line is ( y - y_1=m(x - x_1) ), where ( (x_1,y_1)=(2,-2) ) and ( m = \frac{1}{8} ). Substitute the values into the formula: ( y-(-2)=\frac{1}{8}(x - 2) ). Simplify the equation: [ \begin{align*} y + 2&=\frac{1}{8}x-\frac{2}{8}\ y+2&=\frac{1}{8}x-\frac{1}{4}\ y&=\frac{1}{8}x-\frac{1}{4}-2\ y&=\frac{1}{8}x-\frac{1}{4}-\frac{8}{4}\ y&=\frac{1}{8}x-\frac{9}{4} \end{align*} ]

Answer:

( y=\frac{1}{8}x-\frac{9}{4} )