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

 Algebra Pre-Algebra and Basic Algebra Math Forum

 April 11th, 2011, 04:39 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, 12:29 PM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,155 Thanks: 463 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}$

 Tags 2nd, helpppppp, line, respect, slope

,

# slope of a line with respect to another line

Click on a term to search for related topics.
 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post texaslonghorn15 Calculus 2 June 21st, 2013 08:52 PM ommmmid Algebra 2 May 9th, 2013 01:49 PM jskrzy Algebra 2 October 14th, 2011 10:03 AM MathematicallyObtuse Algebra 4 January 15th, 2011 06:31 AM jessp5798 Calculus 6 February 21st, 2010 10:37 AM

 Contact - Home - Forums - Cryptocurrency Forum - Top