1. Determine an equation of a line that passes through point P that is per pendicular to line m y=-(6)/(7)x-(3)/(7) y=(6)/(7)x-(3)/(7) y=(6)/(7)x+(3)/(7) y=-(6)/(7)x+(3)/(7)

To determine the equation of a line that passes through point P and is perpendicular to line m, we need to find the slope of the perpendicular line.<br /><br />Given that the equation of line m is $y=-\frac{6}{7}x-\frac{3}{7}$, we can see that the slope of line m is $-\frac{6}{7}$.<br /><br />The slope of a line perpendicular to line m is the negative reciprocal of the slope of line m. Therefore, the slope of the perpendicular line is $\frac{7}{6}$.<br /><br />Now, we need to find the equation of the line with slope $\frac{7}{6}$ that passes through point P. Let's assume point P has coordinates $(x_1, y_1)$.<br /><br />Using the point-slope form of a linear equation, we have:<br /><br />$y - y_1 = \frac{7}{6}(x - x_1)$<br /><br />Simplifying this equation, we get:<br /><br />$y = \frac{7}{6}x - \frac{7}{6}x_1 + y_1$<br /><br />Comparing this equation with the given options, we can see that the correct equation of the line is:<br /><br />$y = \frac{7}{6}x - \frac{3}{7}$<br /><br />Therefore, the correct answer is: $y = \frac{7}{6}x - \frac{3}{7}$.