# The equation of a straight line problems

1. Dec 2, 2006

### Equilibrium

1. The problem statement, all variables and given/known data
Find the equations of two lines through A. one parallel and the other perpendicular to the line defined by the given equation.
$$A(-1,1),y=1$$

2. Relevant equations
Parallel lines
$$y = mx + b$$
then solve for b
Perpendicular
$$y = -mx + b[tex] then solve for b 3. The attempt at a solution Parallel: [tex]y=1$$
$$y = 0 + b$$
$$1 = 0 + b$$
$$b = 1[tex] thus, [tex]y=1$$

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

2. Dec 2, 2006

### 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.

3. Dec 2, 2006

### HallsofIvy

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

You write:
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 in to formulas, think! What does the line y= 1 look like? What would a line perpendicular to it look like?

4. Dec 2, 2006

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

5. Dec 2, 2006

### Equilibrium

the slope will be -1?

but x is 0 in y = 0x + 1? could it be that the slope cant be change?

6. Dec 2, 2006