The equation of a straight line problems

In summary, the homework statement says to find the equations of two lines through (-1, 1), one parallel and the other perpendicular to the line defined by the given equation. The parallel lines have slope m and the perpendicular line has slope -1/m. For homework equation y= mx+ b, when y= 1, m= 0, so the equation has slope 0 and for equation y= -mx+ b, when y= 1, b= 1, so the equation has slope 1.
  • #1
Equilibrium
82
0

Homework Statement


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


Homework Equations


Parallel lines
[tex]y = mx + b[/tex]
then solve for b
Perpendicular
[tex]y = -mx + b[tex]
then solve for b

The Attempt at a Solution



Parallel:
[tex]y=1[/tex]
[tex]y = 0 + b[/tex]
[tex]1 = 0 + b[/tex]
[tex]b = 1[tex]
thus,
[tex]y=1[/tex]


Perpendicular:
I got stuck at the perpendicular part... because the answer from the back of my book is x = -1

My answer is y = 1 also
 
Physics news on Phys.org
  • #2
To get a perpendicular line, you need to swap the slopes (m) for their negative recipricol.

Example:

[tex]y=2x+5[/tex]

is perpendicular to [tex]y=-\frac{1}{2}x+5[/tex]

parallel you just change the b.

I don't know if this helps, I hope it does.
 
  • #3
Equilibrium said:

Homework Statement


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


Homework Equations


Parallel lines
[tex]y = mx + b[/tex]
then solve for b
Perpendicular
[tex]y = -mx + b[tex]
then solve for b

The Attempt at a Solution



Parallel:
[tex]y=1[/tex]
[tex]y = 0 + b[/tex]
[tex]1 = 0 + b[/tex]
[tex]b = 1[tex]
thus,
[tex]y=1[/tex]


Perpendicular:
I got stuck at the perpendicular part... because the answer from the back of my book is x = -1

My answer is y = 1 also
?? do you mean you get y= 1 for a parallel and a perpendicular? Surely that can't be possible!


You write:
Parallel lines
[tex]y = mx + b[/tex]
then solve for b
Perpendicular
[tex]y = -mx + b[tex]
then solve for b
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?

Rather than trying to plug numbers into formulas, think! What does the line y= 1 look like? What would a line perpendicular to it look like?
 
  • #4
y = 1 is a horizontal line with slope 0. the perpendicular line will be vertical. what does that tell you about the slope?
 
  • #5
the slope will be -1?

but x is 0 in y = 0x + 1? could it be that the slope can't be change?
 
  • #6
the slope of a vertical line is undefined.
 
  • #7
Equilibrium said:
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?
 

Related to The equation of a straight line problems

1. What is the equation of a straight line?

The equation of a straight line is represented in the form y = mx + b, where m is the slope of the line and b is the y-intercept. This equation describes the relationship between the x and y coordinates of all the points on the line.

2. How do you find the slope of a straight line?

The slope of a straight line can be found by dividing the change in y-coordinates by the change in x-coordinates between any two points on the line. This can also be written as rise over run, or (y2 - y1) / (x2 - x1).

3. What does the y-intercept represent in the equation of a straight line?

The y-intercept, represented by the term b in the equation y = mx + b, is the point where the line crosses the y-axis. It is the value of y when x is equal to 0.

4. Can a straight line have a negative slope?

Yes, a straight line can have a negative slope. This means that the line is decreasing as it moves from left to right, and the value of y decreases as the value of x increases.

5. How do you graph a straight line using its equation?

To graph a straight line using its equation, you can plot the y-intercept on the y-axis and then use the slope to determine other points on the line. You can also plot two points on the line and then use a ruler to draw a straight line through them.

Similar threads

  • Precalculus Mathematics Homework Help
Replies
17
Views
1K
  • Precalculus Mathematics Homework Help
Replies
5
Views
990
  • Precalculus Mathematics Homework Help
Replies
4
Views
806
  • Precalculus Mathematics Homework Help
Replies
14
Views
511
  • Precalculus Mathematics Homework Help
Replies
2
Views
971
  • Precalculus Mathematics Homework Help
Replies
2
Views
577
  • Precalculus Mathematics Homework Help
Replies
7
Views
958
  • Precalculus Mathematics Homework Help
Replies
2
Views
1K
  • Precalculus Mathematics Homework Help
Replies
2
Views
945
  • Precalculus Mathematics Homework Help
Replies
9
Views
2K
Back
Top