EXPLANATION:
Given;
We are given the following linear equation;
[tex]x=-6y-18[/tex]Required;
We are required to write the equation of a line perpendicular to the one given and which passes through the point
[tex]P(0,0)[/tex]Step-by-step solution;
We shall begin by expressing the equation given in slope-intercept form as follows;
[tex]\begin{gathered} Slope-intercept\text{ }form: \\ y=mx+b \end{gathered}[/tex]We shall now make y the subject of the given equation. This is shown below;
[tex]\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}[/tex]We can now simplify;
[tex]6y=-x-18[/tex]Next we divide both sides by 6;
[tex]\frac{6y}{6}=-\frac{x}{6}-\frac{18}{6}[/tex][tex]y=-\frac{1}{6}x-3[/tex]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
[tex]m=-\frac{1}{6}[/tex]The inverse of that would be
[tex]inverse=-\frac{6}{1}[/tex]and the negative of that would be
[tex]Negative\text{ }inverse=\frac{6}{1}=6[/tex]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;
[tex]y=mx+b[/tex]Where we have;
[tex]\begin{gathered} (x,y)=(0,0) \\ m=6 \end{gathered}[/tex]We can have the following;
[tex]\begin{gathered} y=mx+b \\ \Rightarrow0=6(0)+b \end{gathered}[/tex]Simplify this and we'll have;
[tex]0=b[/tex]Now we have the values of m and b.
The equation after substituting for the values of m and b would now be;
ANSWER:
[tex]\begin{gathered} y=6x+0 \\ Therefore: \\ y=6x \end{gathered}[/tex]