What is the equation for a line perpendicular to \(6x – 7y = 8\) and passing through the point (8, 1)?
Explanation
First rearrange the equation into slope-intercept form:
\(y=\frac{6}{7} x-\frac{8}{7}\)
The slope of the perpendicular line is \(\frac{-7}{6}\)
Substituting the coordinates of the point,
\(y=\frac{-7}{6} x+b\)
\(\Leftrightarrow 1=\frac{-7}{6}(8)+b\)
\(\rightarrow \mathbf{b}=\frac{31}{3}\)
→ The equation becomes: \(y=\frac{-7}{6} x+\frac{31}{3}\)