Questions: Consider the line 6x-8y=-8. Find the equation of the line that is parallel to this line and passes through the point (5,-6). Find the equation of the line that is perpendicular to this line and passes through the point (5,-6).

Solution Steps

The slope of the line parallel to the given line is the same, \(m = 0.75\). Using the point-slope form, \(y - y_1 = m(x - x_1)\), with \((x_1, y_1) = (5, -6)", we get \(y + 6 = 0.75(x - 5)\). Rearranging to slope-intercept form, \(y = mx + b'\), we find \(b' = -9.75\). Thus, the equation of the parallel line is \(y = 0.75x - 9.75\).

The slope of the line perpendicular to the given line is the negative reciprocal of \(m\), \(m' = -1/m = -1.33\). Using the point-slope form, \(y - y_1 = m'(x - x_1)\), with \((x_1, y_1) = (5, -6)", we get \(y + 6 = -1.33(x - 5)\). Rearranging to slope-intercept form, \(y = m'x + b''\), we find \(b'' = 0.65\). Thus, the equation of the perpendicular line is \(y = -1.33x + 0.65\).

Final Answer: