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Line pp has an equation of y=−56x−5y = -\frac{5}{6}x - 5. Line qq includes the point (9,−4)(9,-4) and is parallel to line pp. What is the equation of line qq?\newlineWrite the equation in slope-intercept form. Write the numbers in the equation as simplified proper fractions, improper fractions, or integers.

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Q. Line pp has an equation of y=−56x−5y = -\frac{5}{6}x - 5. Line qq includes the point (9,−4)(9,-4) and is parallel to line pp. What is the equation of line qq?\newlineWrite the equation in slope-intercept form. Write the numbers in the equation as simplified proper fractions, improper fractions, or integers.
  1. Determine slope of line p: Determine the slope of line p. The equation of line p is given as y=−56x−5y = -\frac{5}{6}x - 5. The slope-intercept form of a line is y=mx+by = mx + b, where mm is the slope and bb is the y-intercept. By comparing the given equation with the slope-intercept form, we can see that the slope (mm) of line p is −56-\frac{5}{6}.
  2. Find slope of line qq: Since line qq is parallel to line pp, it must have the same slope. Parallel lines have the same slope. Therefore, the slope of line qq will also be −56-\frac{5}{6}.
  3. Use point-slope form: Use the point-slope form to find the equation of line qq. The point-slope form of a line is y−y1=m(x−x1)y - y_1 = m(x - x_1), where mm is the slope and (x1,y1)(x_1, y_1) is a point on the line. We know that line qq passes through the point (9,−4)(9, -4) and has a slope of −56-\frac{5}{6}. Plugging these values into the point-slope form, we get: y−(−4)=−56(x−9)y - (-4) = -\frac{5}{6}(x - 9)
  4. Simplify equation to slope-intercept form: Simplify the equation to get it into slope-intercept form.\newlineFirst, distribute the slope on the right side of the equation:\newliney+4=−56x+(56)⋅9y + 4 = -\frac{5}{6}x + (\frac{5}{6})\cdot9\newlineNow, simplify the right side:\newliney+4=−56x+456y + 4 = -\frac{5}{6}x + \frac{45}{6}\newliney+4=−56x+152y + 4 = -\frac{5}{6}x + \frac{15}{2}\newlineNext, subtract 44 from both sides to solve for yy:\newliney=−56x+152−4y = -\frac{5}{6}x + \frac{15}{2} - 4\newliney=−56x+152−82y = -\frac{5}{6}x + \frac{15}{2} - \frac{8}{2}\newliney=−56x+72y = -\frac{5}{6}x + \frac{7}{2}

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