Question

Onsider the line y=-5+6.

find the equation of the line perpendicular to this line and passes through the point (-2,-3)
find the equation of the is parallel to this line and passes through the point (-2,-3)​

Answer 1

The equation of the line perpendicular to the given line is

`y=\frac 1 5 x + \frac {-13}{5}`.

The equation of the line perpendicular to the given line is

`y=-5 x -13`.

Explanation:

The given line is y=-5x+6

The slope of the given line,
`m_1=-5`

Let
`y= m_2 x + C_2` be the slope of the line perpendicular to y=-5x+6,

where
`C_2` is constant.

As multiplication of slopes of two perpendicular lines equal to -1, so

`m_1m_2=-1 \\\\\Rightarrow -5* m_2= -1 \\\\\Rightarrow m_2= \frac {-1}{-5}= \frac 1 5`

So, the equation of the line perpendicular to the given line is

`y=\frac 1 5 x +C_2` which passes through the point (-2,-3), so

`-3=\frac 1 5 (-2) +C_2 \\\\\Rightarrow C_2 = -3 +\frac 2 5= \frac {-13}{5}`

Hence, the equation of the line perpendicular to the given line is

`y=\frac 1 5 x + \frac {-13}{5}`.

Now, let
`y=m_3 x +C_3` be the equation of the parallel line to y=-5x+6.

As the slopes of two parallel lines are the same, so
`m_3=-5`.

So, the equation of the parallel line becomes

`y=-5 x +C_3` which passes through the point (-2,-3), so

`-3=-5 (-2) +C_3 \\\\\Rightarrow C_2 = -3 - 10 = -13`

Hence, the equation of the line perpendicular to the given line is

`y=-5 x -13`.