9. write the equation of the line that is parallel to the line 5x - 4y = 4 and passes through the point (-8…

9. write the equation of the line that is parallel to the line 5x - 4y = 4 and passes through the point (-8, 2).\n10. write the equation of the line that is perpendicular to the line y = -1/5x + 9 and passes through the point (-2, -2).\n11. write the equation of the line that is perpendicular to the line 5x + 6y = 18 and passes through the point (10, 7).\n12. write the equation of the line that is perpendicular to the line x - 4y = 20 and passes through the point (2, -5).\n13. cd is formed by c(-5, 9) and d(7, 5). if line t is the perpendicular bisector of cd, write a linear equation for t in slope - intercept form.\n14. pq is formed by p(10, 4) and q(2, -8). if line k is the perpendicular bisector of pq, write a linear equation for k in slope - intercept form.

9. write the equation of the line that is parallel to the line 5x - 4y = 4 and passes through the point (-8, 2).\n10. write the equation of the line that is perpendicular to the line y = -1/5x + 9 and passes through the point (-2, -2).\n11. write the equation of the line that is perpendicular to the line 5x + 6y = 18 and passes through the point (10, 7).\n12. write the equation of the line that is perpendicular to the line x - 4y = 20 and passes through the point (2, -5).\n13. cd is formed by c(-5, 9) and d(7, 5). if line t is the perpendicular bisector of cd, write a linear equation for t in slope - intercept form.\n14. pq is formed by p(10, 4) and q(2, -8). if line k is the perpendicular bisector of pq, write a linear equation for k in slope - intercept form.

Answer

Explanation:

Step1: Find slope of parallel or perpendicular line

For parallel lines, slopes are equal. For perpendicular lines, the product of slopes is - 1. For the line (5x - 4y=4), rewrite it in slope - intercept form (y = mx + b) ((m) is slope, (b) is y - intercept). [ \begin{align*} 5x-4y&=4\ -4y&=-5x + 4\ y&=\frac{5}{4}x-1 \end{align*} ] The slope of the line parallel to it is (m_1=\frac{5}{4}). Using the point - slope form (y - y_1=m(x - x_1)) with ((x_1,y_1)=(-8,2)), we have (y - 2=\frac{5}{4}(x + 8)). [ \begin{align*} y-2&=\frac{5}{4}x+10\ y&=\frac{5}{4}x + 12 \end{align*} ] For the line (y=-\frac{1}{5}x + 9), the slope of the perpendicular line is (m_2 = 5) (since (-\frac{1}{5}\times m_2=-1)). Using the point - slope form with ((x_1,y_1)=(-2,-2)), we get (y+2 = 5(x + 2)). [ \begin{align*} y+2&=5x+10\ y&=5x + 8 \end{align*} ] For the line (5x + 6y=18), rewrite it in slope - intercept form: [ \begin{align*} 6y&=-5x + 18\ y&=-\frac{5}{6}x+3 \end{align*} ] The slope of the perpendicular line is (m_3=\frac{6}{5}). Using the point - slope form with ((x_1,y_1)=(10,7)), we have (y - 7=\frac{6}{5}(x - 10)). [ \begin{align*} y-7&=\frac{6}{5}x-12\ y&=\frac{6}{5}x-5 \end{align*} ] For the line (x-4y = 20), rewrite it in slope - intercept form: [ \begin{align*} -4y&=-x + 20\ y&=\frac{1}{4}x-5 \end{align*} ] The slope of the perpendicular line is (m_4=-4). Using the point - slope form with ((x_1,y_1)=(2,-5)), we get (y + 5=-4(x - 2)). [ \begin{align*} y+5&=-4x + 8\ y&=-4x+3 \end{align*} ] For the line segment (\overline{CD}) with (C(-5,9)) and (D(7,5)), the slope of (\overline{CD}) is (m_{CD}=\frac{5 - 9}{7+5}=\frac{-4}{12}=-\frac{1}{3}). The mid - point of (\overline{CD}) is ((\frac{-5 + 7}{2},\frac{9+5}{2})=(1,7)). The slope of the perpendicular bisector (t) is (m_t = 3). Using the point - slope form, (y - 7=3(x - 1)), so (y=3x+4). For the line segment (\overline{PQ}) with (P(10,4)) and (Q(2,-8)), the slope of (\overline{PQ}) is (m_{PQ}=\frac{-8 - 4}{2 - 10}=\frac{-12}{-8}=\frac{3}{2}). The mid - point of (\overline{PQ}) is ((\frac{10 + 2}{2},\frac{4-8}{2})=(6,-2)). The slope of the perpendicular bisector (k) is (m_k=-\frac{2}{3}). Using the point - slope form, (y + 2=-\frac{2}{3}(x - 6)), so (y=-\frac{2}{3}x+2).

Answer:

  1. (y=\frac{5}{4}x + 12)
  2. (y=5x + 8)
  3. (y=\frac{6}{5}x-5)
  4. (y=-4x+3)
  5. (y=3x+4)
  6. (y=-\frac{2}{3}x+2)