Question

Consider the line y=9x-3.Find the equation of the line that is perpendicular to this line and passes through the point (-8, – 5).

Answer 1

As per given by the question,

There are given that the equation of line is

`y=9x-3`

And, the point is (-8, -5).

Now,

Let us find the slope of line perpendicular to given line.

Then,

According to the properties of perpendicular,

We know that the product of slopes of the given line and slope of line perpendicular to given line is equal to -1.

So,

9x slope of line perpendicular to given line =-1.

Then,

the slope of line is

`-(1)/(9)`

Now,

Find the equation of line with slope -1/9 and passes through the point (-8, -5).

So,

Substitute the value of slope(m) and point (-8, -5) in the general form of equation line, y=mx+c

Then,

`\begin{gathered} y=mx+c \\ -5=-(1)/(9)(-8)+c \\ -5=(8)/(9)+c \\ 8+9c=-45 \\ 9c=-45-8 \\ 9c=-53 \\ c=(-53)/(9) \end{gathered}`

Then,

Put the value of m and c into the above equation

So,

`\begin{gathered} y=mx+c \\ y=-(1)/(9)x-(53)/(9) \\ y=-(x)/(9)-(53)/(9) \end{gathered}`

Hence, the equation of line is shown below:

`y=-(x)/(9)-(53)/(9)`