What is the equation for a line perpendicular to \(6x – 7y = 8\) and passing through the point (8, 1)?

\(y=\frac{-7}{6} x+\frac{31}{3}\)


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: \(y=\frac{-7}{6} x+\frac{31}{3}\)

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