My Math Forum slope of line in respect to 2nd line helpppppp please

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 April 11th, 2011, 03:39 AM #1 Newbie   Joined: Apr 2011 Posts: 8 Thanks: 0 slope of line in respect to 2nd line helpppppp please 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! this is another example i have to study for this test tomorrow havent a clue how to do this please explain thank you
 April 11th, 2011, 11:29 AM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,164 Thanks: 472 Math Focus: Calculus/ODEs Re: slope of line in respect to 2nd line helpppppp please a) Let $m_1$ and $m_2$ represent the slopes of the given lines $L_1$ and $L_2$ respectively. Using the fact that: $m_n=\tan$$\theta_n$$$ we can find the respective slope m between the two lines by taking the difference of the angles: $m=\tan$$\tan^{-1}\(m_1$$-\tan^{-1}$$m_2$$\)=\frac{\tan$$\tan^{-1}\(m_1$$\)-\tan$$\tan^{-1}\(m_2$$\)}{1+\tan$$\tan^{-1}\(m_1$$\)\tan$$\tan^{-1}\(m_2$$\)}$ $m=\frac{m_1-m_2}{1+m_1m_2}$ For this problem, we have $m_1=2,m_2=1$ thus: $m=\frac{2-1}{1+2\cdot1}=\frac{1}{3}$ b) Using rotation of axes, where the angle of rotation is $\theta=\tan^{-1}$$m_2$$$ we have: $\cos\theta=\frac{1}{\sqrt{m_2^2+1}}$ and $\sin\theta=\frac{m_2}{\sqrt{m_2^2+1}}$ We use: $x=X\cos\theta-Y\sin\theta=X\frac{1}{\sqrt{m_2^2+1}}-Y\frac{m_2}{\sqrt{m_2^2+1}}=\frac{1}{\sqrt{m_2^2+1 }}$$X-m_2Y$$$ $y=X\sin\theta+Y\cos\theta=X\frac{m_2}{\sqrt{m_2^2+ 1}}+Y\frac{1}{\sqrt{m_2^2+1}}=\frac{1}{\sqrt{m_2^2 +1}}$$m_2X+Y$$$ In the new coordinate system, the line $y=m_1x+b$ becomes: $\frac{1}{\sqrt{m_2^2+1}}$$m_2X+Y$$=m_1$$\frac{1}{\ sqrt{m_2^2+1}}\(X-m_2Y$$\)+b$ $m_2X+Y=m_1X-m_1m_2Y+b\sqrt{m_2^2+1}$ $Y+m_1m_2Y=m_1X-m_2X+b\sqrt{m_2^2+1}$ $Y$$1+m_1m_2$$=$$m_1-m_2$$X+b\sqrt{m_2^2+1}$ $Y=\frac{$$m_1-m_2$$X+b\sqrt{m_2^2+1}}{1+m_1m_2}$ Thus, we find with $m_1=2,m_2=1,b=1$ we have the line: $Y=\frac{$$2-1$$X+\sqrt{1^2+1}}{1+2\cdot1}=\frac{1}{3}X+\frac{\ sqrt{2}}{3}$

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# slope of a line with respect to another line

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