1. **Problem:** Find the equation of the line that passes through the point (6, -7) and is parallel to the line $$2x - 3y + 4 = 0$$. 2. **Formula and rules:** - The slope-intercept form of a line is $$y = mx + c$$ where $$m$$ is the slope. - Lines are parallel if they have the same slope. - To find the slope of the given line, rewrite it in slope-intercept form. 3. **Find the slope of the given line:** $$2x - 3y + 4 = 0$$ Rearranged: $$-3y = -2x - 4$$ Divide both sides by $$-3$$: $$y = \frac{\cancel{-2}}{\cancel{-3}}x + \frac{\cancel{-4}}{\cancel{-3}} = \frac{2}{3}x + \frac{4}{3}$$ So, the slope $$m = \frac{2}{3}$$. 4. **Slope of the parallel line:** Since parallel lines have the same slope, the new line's slope is also $$m = \frac{2}{3}$$. 5. **Use point-slope form to find the equation:** Point-slope form: $$y - y_1 = m(x - x_1)$$ Substitute $$m = \frac{2}{3}$$ and point $$(6, -7)$$: $$y - (-7) = \frac{2}{3}(x - 6)$$ $$y + 7 = \frac{2}{3}x - 4$$ 6. **Simplify to slope-intercept form:** $$y = \frac{2}{3}x - 4 - 7$$ $$y = \frac{2}{3}x - 11$$ **Final answer:** $$\boxed{y = \frac{2}{3}x - 11}$$