You are given the line l with equation [tex]y=\dfrac{2}{3}x-4.[/tex]
1. The equation of line that passes through the point (-2,-5) and is parallel to the line l.
Parallel lines have the same slope. So the slope of unknown line is [tex]\dfrac{2}{3}.[/tex]
Then the equation is
[tex]y=\dfrac{2}{3}x+a.[/tex]
This line passes through point (-2,-5), this means that coordinates of this point satisfy the equation, substitute x=-2 and y=-5 into equation:
[tex]-5=\dfrac{2}{3}\cdot (-2)+a,\\ \\a=-5+\dfrac{4}{3}=\dfrac{-15+4}{3}=-\dfrac{11}{3}.[/tex]
Thus, the equation of parallel line is
[tex]y=\dfrac{2}{3}x-\dfrac{11}{3}.[/tex]
2. The equation of line that passes through the point (-2,-5) and is perpendicular to the line l.
Perpendicular lines have slopes that satisfy the condition
[tex]m_1\cdot m_2=-1.[/tex]
Therefore, the slope of perpendicular line is
[tex]\dfrac{2}{3}\cdot m_2=-1,\\ \\m_2=-\dfrac{3}{2}.[/tex]
Then the equation is
[tex]y=-\dfrac{3}{2}x+b.[/tex]
This line passes through point (-2,-5), this means that coordinates of this point satisfy the equation, substitute x=-2 and y=-5 into equation:
[tex]-5=-\dfrac{3}{2}\cdot (-2)+b,\\ \\b=-5-3=-8.[/tex]
Thus, the equation of perpendicular line is
[tex]y=-\dfrac{3}{2}x-8.[/tex]
