Consider the line y=-(3)/(7)x-6 Find the equation of the line that is perpendicular to this line and passe the point (-7,5) Find the equation of the line that is parallel to this line and passes thro point (-7,5) Equation of perpendicular line: square Equation of parallel line: square

To find the equation of the line that is perpendicular to the given line and passes through the point $$(-7,5)$$ , we need to determine the slope of the perpendicular line.

The slope of the given line is $$-\frac{3}{7}$$ . The slope of a line perpendicular to this line is the negative reciprocal of $$-\frac{3}{7}$$ , which is $$\frac{7}{3}$$ .

Using the point-slope form of a linear equation, we can write the equation of the perpendicular line as:

$$y - 5 = \frac{7}{3}(x + 7)$$

Simplifying this equation, we get:

$$y = \frac{7}{3}x + \frac{49}{3}$$

Therefore, the equation of the perpendicular line is $$y = \frac{7}{3}x + \frac{49}{3}$$ .

To find the equation of the line that is parallel to the given line and passes through the point $$(-7,5)$$ , we can use the same slope as the given line, which is $$-\frac{3}{7}$$ .

Using the point-slope form of a linear equation, we can write the equation of the parallel line as:

$$y - 5 = -\frac{3}{7}(x + 7)$$

Simplifying this equation, we get:

$$y = -\frac{3}{7}x - \frac{6}{7}$$

Therefore, the equation of the parallel line is $$y = -\frac{3}{7}x - \frac{6}{7}$$ .

In summary:
Equation of perpendicular line: $$y = \frac{7}{3}x + \frac{49}{3}$$
Equation of parallel line: $$y = -\frac{3}{7}x - \frac{6}{7}$$