Given the linear equation;
[tex]y=-x-5[/tex]Note that the slope of a line perpendicular to another has its slope as a negative inverse of the first one.
This means if the slope of line 1 is m, then the slope of a perpendicular line which is line 2 would be the negative inverse of m, that is;
[tex]\begin{gathered} \text{Slope}1=m \\ \text{Slope}2=-\frac{1}{m} \end{gathered}[/tex]The slope of the line given is -1. The slope is the coefficient of x.
For a line perpendicular to this one;
[tex]\begin{gathered} \text{Slope}1=-1 \\ \text{Slope}2=-(\frac{1}{-1}) \\ \text{Slope}2=\frac{1}{1} \\ \text{Slope}2=1 \end{gathered}[/tex]The slope of the perpendicular line is 1, and we already have a point along this line given as
[tex](9,0)[/tex]The equation of a line in point-slope form is;
[tex]y-y_1=m(x-x_1)[/tex]Where the variables are;
[tex]x_1=9,y_1=0,m=1[/tex]The equation in point-slope form now becomes;
[tex]y-0=1(x-9)[/tex]For the equation in slope-intercept form, the general form of the equation is;
[tex]y=mx+b[/tex]Here also the variables are;
[tex]x=9,y=0,m=1[/tex]Substituting the variables we now have;
[tex]\begin{gathered} y=mx+b \\ 0=1(9)+b \\ 0=9+b \\ \text{Subtract 9 from both sides;} \\ b=-9 \end{gathered}[/tex]The equation now would be re-written as;
[tex]\begin{gathered} y=mx+b \\ y=1x+(-9) \\ y=x-9 \end{gathered}[/tex]ANSWER:
[tex]\begin{gathered} \text{ Point-slope form:} \\ y-\lbrack0\rbrack=\lbrack1\rbrack(x-\lbrack9\rbrack) \end{gathered}[/tex][tex]\begin{gathered} \text{Slope}-\text{intercept form:} \\ y=\lbrack1\rbrack x+\lbrack-9\rbrack \end{gathered}[/tex]