Question

Write an equation of the line that passes through the given point and is perpendicular to the given line. Your answer should be written in slope-intercept form.P(0, 0), x = −6y − 18

Answer 1

Step-by-step explanation:

Given;

We are given the following linear equation;

`x=-6y-18`

Required;

We are required to write the equation of a line perpendicular to the one given and which passes through the point

`P(0,0)`

Step-by-step solution;

We shall begin by expressing the equation given in slope-intercept form as follows;

`\begin{gathered} Slope-intercept\text{ }form: \\ y=mx+b \end{gathered}`

We shall now make y the subject of the given equation. This is shown below;

`\begin{gathered} x=-6y-18 \\ Add\text{ }6y\text{ }to\text{ }both\text{ }sides: \\ x+6y=-6y+6y-18 \\ Subtract\text{ }x\text{ }from\text{ }both\text{ }sides: \\ x-x+6y=-6y+6y-x-18 \end{gathered}`

We can now simplify;

`6y=-x-18`

Next we divide both sides by 6;

`(6y)/(6)=-(x)/(6)-(18)/(6)`
`y=-(1)/(6)x-3`

We now have the equation in the 'slope-intercept' form.

Take note that the coefficient of x is the slope of the line. Also, note that for a line perpendicular to another one, the slope of one would be a negative inverse of the other.

Therefore, we have the slope of this line as

`m=-(1)/(6)`

The inverse of that would be

`inverse=-(6)/(1)`

and the negative of that would be

`Negative\text{ }inverse=(6)/(1)=6`

Therefore, we need to find the equation whose slope is 6, and passes through the point (0, 0).

Using the general form of the equation;

`y=mx+b`

Where we have;

`\begin{gathered} (x,y)=(0,0) \\ m=6 \end{gathered}`

We can have the following;

`\begin{gathered} y=mx+b \\ \Rightarrow0=6(0)+b \end{gathered}`

Simplify this and we'll have;

`0=b`

Now we have the values of m and b.

The equation after substituting for the values of m and b would now be;

ANSWER:

`\begin{gathered} y=6x+0 \\ Therefore: \\ y=6x \end{gathered}`