slope in respect to an equation help

• Apr 8th 2011, 02:08 PM
jessmari86
slope in respect to an equation help
Consider two straight lines L1 (y = 2x +1) and L2 (y=x).

a) What is the slope of line L1 in respect to line L2?

b) What is the equation of line L1 in respect to line L2?

NOTE. Think that the original xy co-ordinate system has been

rotated in respect to the origin so that line L2 defines the new

x-axis x´ and y´ is the new y-axis!

now i know the slope of line 1 is 2 and line 2 is 1. im not sure if thats the kind of answer they want for part a of the question. and part b im not sure at all what they are asking for. PLease help me understand this sort of problem.

thank you
• Apr 8th 2011, 09:05 PM
TKHunny
Not sure your in the right classification. The most useful formulation would be in the expression for the tangent of the difference of two angles.
• Apr 8th 2011, 09:07 PM
Sambit
You are rotating the axes by an angle $\frac{\pi}{4}$. Let us denote this angle by $\theta$. Further, let the given equation of line 1 is $f(x,y)=0$. Then the transformed equation of line 1 after the rotation is given by: $f( x'\cos\theta-y' \sin\theta, x' \sin\theta + y' \cos\theta) = 0$ where the new axes are $x'=0 , y'=0$.
• Apr 9th 2011, 12:29 AM
jessmari86
Quote:

Originally Posted by Sambit
You are rotating the axes by an angle $\frac{\pi}{4}$. Let us denote this angle by $\theta$. Further, let the given equation of line 1 is $f(x,y)=0$. Then the transformed equation of line 1 after the rotation is given by: $f( x'\cos\theta-y' \sin\theta, x' \sin\theta + y' \cos\theta) = 0$ where the new axes are $x'=0 , y'=0$.

ok from what i get is i first of course graph the 2 lines the first obvioulsy intersecting at 1,0 on the x axis and the 2nd at 0,0. i can conclude that they intersect at point (-1,-1). Now the only thing i do not understand is the rotation and how you came up with $f( x'\cos\theta-y' \sin\theta, x' \sin\theta + y' \cos\theta) = 0$. what theorum or rule is that called? i do not rememebr learning this so this is new to me. but of course it works to get the new (0,0) point for the new axis by plugging in -1 for x and y and using $\frac{\pi}{4}$. just i guess the only thing is like i said i need to know why to use the formula to get it if you could tell me that would be appreciated
• Apr 9th 2011, 02:57 AM
Sambit
The matrix $R(\theta) = \begin{bmatrix}
\cos \theta & -\sin \theta \\
\sin \theta & \cos \theta \\
\end{bmatrix}$
is the rotation matrix in 2-dimensional form. It gives you the relation $\begin{bmatrix}
x' \\
y' \\
\end{bmatrix} = \begin{bmatrix}
\cos \theta & -\sin \theta \\
\sin \theta & \cos \theta \\
\end{bmatrix}\begin{bmatrix}
x \\
y \\
\end{bmatrix}$
. Hence the formula. Then you just replace the current co-ordinates with the former one.
• Apr 9th 2011, 07:06 AM
HallsofIvy
Finding the angle as you rotate y= x to the x-axis is exactly the same as finding the angle between the lines themselves.

I would use the dot product: The vector <1, 1> is in the direction of the line y= x. The vector <1, 2> is in the direction of the line y= 2x+ 1. Their dot product is $<1, 1>\cdot<1, 2>= 1+ 2= 3= |<1, 1>||<1, 2>|cos(\theta)$ where $\theta$ is the angle between the lines. $|<1, 1>|= \sqrt{1+ 1}= \sqrt{2}$ and $|<1, 2>|= \sqrt{1+ 4}= \sqrt{5}$ so, finally, $cos(\theta)= \frac{3}{\sqrt{10}}$.

You could use a calculator to find the angle and then the tangent but you could also say $tan(\theta)= \frac{sin(\theta)}{cos(\theta)}$ $= \frac{\sqrt{1- cos^2(\theta)}}{cos(\theta)}$ $= \frac{\sqrt{1- \frac{9}{10}}}{\frac{3}{\sqrt{10}}}$ $= \frac{1}{3}$.

Perhaps simplest, though, is to use the formula $tan(x- y)= \frac{tan(x)- tan(y)}{1+ tan(x)(tan)y)}$. Here, tan(x)= 2 and tan(y)= 1 so $tan(x- y)= \frac{2- 1}{1+ 2(1)}= \frac{1}{3}$!