Answer:
The equation of the line perpendicular to the given line is
[tex]y=\frac 1 5 x + \frac {-13}{5}[/tex].
The equation of the line perpendicular to the given line is
[tex]y=-5 x -13[/tex].
Step-by-step explanation:
The given line is y=-5x+6
The slope of the given line, [tex]m_1=-5[/tex]
Let [tex]y= m_2 x + C_2[/tex] be the slope of the line perpendicular to y=-5x+6,
where [tex]C_2[/tex] is constant.
As multiplication of slopes of two perpendicular lines equal to -1, so
[tex]m_1m_2=-1 \\\\\Rightarrow -5\times m_2= -1 \\\\\Rightarrow m_2= \frac {-1}{-5}= \frac 1 5[/tex]
So, the equation of the line perpendicular to the given line is
[tex]y=\frac 1 5 x +C_2[/tex] which passes through the point (-2,-3), so
[tex]-3=\frac 1 5 (-2) +C_2 \\\\\Rightarrow C_2 = -3 +\frac 2 5= \frac {-13}{5}[/tex]
Hence, the equation of the line perpendicular to the given line is
[tex]y=\frac 1 5 x + \frac {-13}{5}[/tex].
Now, let [tex]y=m_3 x +C_3[/tex] be the equation of the parallel line to y=-5x+6.
As the slopes of two parallel lines are the same, so [tex]m_3=-5[/tex].
So, the equation of the parallel line becomes
[tex]y=-5 x +C_3[/tex] which passes through the point (-2,-3), so
[tex]-3=-5 (-2) +C_3 \\\\\Rightarrow C_2 = -3 - 10 = -13[/tex]
Hence, the equation of the line perpendicular to the given line is
[tex]y=-5 x -13[/tex].
