The equation of a straight line problems

[Equilibrium]

Homework Statement

Find the equations of two lines through A. one parallel and the other perpendicular to the line defined by the given equation.
$$A(-1,1),y=1$$

Homework Equations

Parallel lines
$$y = mx + b$$
then solve for b
Perpendicular
[tex]y = -mx + b[tex] then solve for b<br /> <br /> <h2>The Attempt at a Solution</h2><br /> <br /> Parallel:<br /> $$y=1$$<br /> $$y = 0 + b$$<br /> $$1 = 0 + b$$<br /> [tex]b = 1[tex] thus,<br /> $$y=1$$<br /> <br /> <br /> Perpendicular:<br /> I got stuck at the perpendicular part... because the answer from the back of my book is x = -1<br /> <br /> My answer is y = 1 also[/tex][/tex][/tex][/tex]

 

[dontdisturbmycircles]

To get a perpendicular line, you need to swap the slopes (m) for their negative recipricol.

Example:

$$y=2x+5$$

is perpendicular to $$y=-\frac{1}{2}x+5$$

parallel you just change the b.

I don't know if this helps, I hope it does.

 

[HallsofIvy]

Homework Statement

Find the equations of two lines through A. one parallel and the other perpendicular to the line defined by the given equation.
$$A(-1,1),y=1$$

Homework Equations

Parallel lines
$$y = mx + b$$
then solve for b
Perpendicular
[tex]y = -mx + b[tex] then solve for b<br /> <br /> <h2>The Attempt at a Solution</h2><br /> <br /> Parallel:<br /> $$y=1$$<br /> $$y = 0 + b$$<br /> $$1 = 0 + b$$<br /> [tex]b = 1[tex] thus,<br /> $$y=1$$<br /> <br /> <br /> Perpendicular:<br /> I got stuck at the perpendicular part... because the answer from the back of my book is x = -1<br /> <br /> My answer is y = 1 also[/tex][/tex][/tex][/tex]

[tex][tex][tex][tex] ?? do you mean you get y= 1 for a parallel and a perpendicular? Surely that can't be possible!<br /> <br /> <br /> You write:<br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> Parallel lines<br /> $$y = mx + b$$<br /> then solve for b<br /> Perpendicular<br /> [tex]y = -mx + b$$then solve for b$$[/tex] </div> </div> </blockquote>[tex]$$That's incorrect. A line perependicular to y= mx+ b has slope -1/m, not -m. Since for y= 1, m= 0, -1/m does NOT exist. What does that tell you?<br /> <br /> Rather than trying to plug numbers into formulas, <b>think</b>! What does the line y= 1 look like? What would a line perpendicular to it look like?$$[/tex][/tex][/tex][/tex][/tex]

 

[courtrigrad]

y = 1 is a horizontal line with slope 0. the perpendicular line will be vertical. what does that tell you about the slope?

 

[Equilibrium]

the slope will be -1?

but x is 0 in y = 0x + 1? could it be that the slope can't be change?

 

[courtrigrad]

the slope of a vertical line is undefined.

 

[HallsofIvy]

the slope will be -1?

but x is 0 in y = 0x + 1? could it be that the slope can't be change?

No, x is NOT 0 in "y= 0x+ 1"! The COEFFICIENT of x is 0. x can be any value.

As Courtigrad pointed out, a vertical line does not have a slope. That was my point before. In fact, my other point was that you shouldn't be worrying about 'slope' at all! y= 1 is a horizontal line. Any line parallel to it must also be a horizontal line, of the form y= constant. Of course, that means it has slope 0 but that is not really important and confuses the issue with "perpendicular" lines. Vertical lines cannot be written in the form y= mx+ b! Obviously any line perpendicular to a horizontal line is vertical. What does the equation of any vertical line look like? What must the equation of a vertical line through (-1, 1) be?