Question

Help please. Was out due to health trying to catch up and learn my work. Thank you so much in advance!

Answer 1

Given the linear equation;

`y=-x-5`

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;

`\begin{gathered} \text{Slope}1=m \\ \text{Slope}2=-(1)/(m) \end{gathered}`

The slope of the line given is -1. The slope is the coefficient of x.

For a line perpendicular to this one;

`\begin{gathered} \text{Slope}1=-1 \\ \text{Slope}2=-((1)/(-1)) \\ \text{Slope}2=(1)/(1) \\ \text{Slope}2=1 \end{gathered}`

The slope of the perpendicular line is 1, and we already have a point along this line given as

`(9,0)`

The equation of a line in point-slope form is;

`y-y_1=m(x-x_1)`

Where the variables are;

`x_1=9,y_1=0,m=1`

The equation in point-slope form now becomes;

`y-0=1(x-9)`

For the equation in slope-intercept form, the general form of the equation is;

`y=mx+b`

Here also the variables are;

`x=9,y=0,m=1`

Substituting the variables we now have;

`\begin{gathered} y=mx+b \\ 0=1(9)+b \\ 0=9+b \\ \text{Subtract 9 from both sides;} \\ b=-9 \end{gathered}`

The equation now would be re-written as;

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

ANSWER:

`\begin{gathered} \text{ Point-slope form:} \\ y-\lbrack0\rbrack=\lbrack1\rbrack(x-\lbrack9\rbrack) \end{gathered}`
`\begin{gathered} \text{Slope}-\text{intercept form:} \\ y=\lbrack1\rbrack x+\lbrack-9\rbrack \end{gathered}`