Thread: Lines with the same slope.

1. Lines with the same slope.

I have a problem in proving this statement:

If two lines have the same slope then they are either parallel or incidental (the same line).

Thank You!!

2. Re: Lines with the same slope.

y=3x+3 and y=3x-10 are parallel.

but

y=3x+3 and y=3x+3 are the same.

Therefore the difference in these examples is the influence of the constant term.

3. Re: Lines with the same slope.

Suggestion: let $\displaystyle l_1$ have the equation $\displaystyle y=mx+c$
and let $\displaystyle l_2$ have the equation $\displaystyle y=mx+k$. These are two lines which have the same slope.

Now solve for the points of intersection. Parallel lines will never intersect, so we can deduce that the only point of interception is where $\displaystyle c=k$ and the lines are incidental; otherwise, they are parallel.

4. Re: Lines with the same slope.

thank you. i can't view the some parts of your reply though. im new here.

thank you.

6. Re: Lines with the same slope.

Originally Posted by abutazil352
thank you. i can't view the some parts of your reply though. im new here.
I assume you were referring to the use of latex in my post? Where are you having a problem?

7. Re: Lines with the same slope.

Originally Posted by Quacky
I assume you were referring to the use of latex in my post? Where are you having a problem?

Don't you just img.top {vertical-align:15%;}
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math?
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8. Re: Lines with the same slope.

The slope m of a line may be converted to an angle θ using:

$\displaystyle \theta=\tan^{-1}(m)$

Suppose we have two slopes $\displaystyle m_1$ and $\displaystyle m_2$.

If the slopes are parallel or perpendicular, we may state:

$\displaystyle \theta_1-\theta_2=\theta_2-\theta_1$

$\displaystyle \tan^{-1}(m_1)-\tan^{-1}(m_2)=\tan^{-1}(m_2)-\tan^{-1}(m_1)$

$\displaystyle 2\tan^{-1}(m_1)=2\tan^{-1}(m_2)$

Taking the tangent of both sides and applying the double-angle identity for tangent and dividing through by 2, we have:

$\displaystyle \frac{m_1}{1-m_1^2}=\frac{m_2}{1-m_2^2}$

$\displaystyle m_1(1-m_2^2)=m_2(1-m_1^2)$

$\displaystyle m_1-m_1m_2^2-m_2+m_2m_1^2=0$

$\displaystyle (m_1-m_2)(m_1m_2+1)=0$

The root $\displaystyle m_1=m_2$ corresponds to parallel slopes.

The root $\displaystyle m_1m_2=-1$ corresponds to perpendicular slopes.