Transformational geometry: Given the lines L and K, find....

[oshea.emma]

Given the lines L : y=4x and K: x=4y.
f is the transformation (x,y)->(x',y')
where x'=2x-y and y'=x+3y
I'm asked to find the measure of the ACUTE angle between f(L) and f(K) correct to the nearest degree?
:shock:
What will i use to find the angle?
Also what's the point in these types of questions?

 

[soroban]

Re: Transformational geometry question. Genius alert!!

Hello, oshea.emma!

Given the lines \(\displaystyle L:\;y\,=\,4x\) and \(\displaystyle K:\:x\,=\,4y\)

\(\displaystyle f\) is the transformation: \(\displaystyle \,(x,y)\:\rightarrow\x',y')\), where \(\displaystyle x'\:=\:2x\,-\,y\) and \(\displaystyle y'\:=\:x\,+\,3y\)

Find the measure of the acute angle between \(\displaystyle f(L)\) and \(\displaystyle f(K)\) to the nearest degree

Game plan

Find the slopes of \(\displaystyle F(L)\) and \(\displaystyle f(K):\;m_1,\;m_2.\)

\(\displaystyle \;\;\;\;\)then use the formula: \(\displaystyle \,\tan\theta \;=\;\frac{m_2 - m_1}{1\,+\,m_1m_2}\)


On line \(\displaystyle L\), choose two points: \(\displaystyle \,(0,0),\;(1,4)\)

\(\displaystyle \;\;f(0,0) \;=\;(2\cdot0-0,\,0+3\cdot0) \;= \;(0,0)\)
\(\displaystyle \;\;f(1,4)\;=\;(2\cdot1-4,\,1+3\cdot4) \;=\;(-2,13)\)

\(\displaystyle \;\;\)The slope of \(\displaystyle f(L)\) is:\(\displaystyle \,m_1 \:=\:\frac{13\,-\,0}{-2\,-\,0}\:=\:-\frac{13}{2}\)


On line \(\displaystyle K\), choose two points: \(\displaystyle \,(0,0),\;(4,1)\)

\(\displaystyle \;\;f(0,0) \;=\;(0,0)\)
\(\displaystyle \;\;f(4,1)\;=\;2\cdot4-1,\,4=3\cdot1)\;=\;(7,7)\)

\(\displaystyle \;\;\)The slope of \(\displaystyle f(K)\) is: \(\displaystyle \,m_2\;=\:\frac{7\,-\,0}{7\,-\,0} \:=\:1\)

Hence, we have: \(\displaystyle \,\tan\theta \;= \;\L\frac{1\,-\,\left(-\frac{13}{2}\right)}{1\,+\,\left(-\frac{13}{2}\right)(1)}\)\(\displaystyle \;= \;-\frac{15}{11}\)

Since we want the acute angle, we use: \(\displaystyle \,\left|-\frac{15}{11}\right|\:=\:\frac{15}{11}\)

We have: \(\displaystyle \,\tan\theta\:=\:\frac{15}{11}\)

\(\displaystyle \;\;\)Therefore: \(\displaystyle \,\theta\:=\:\tan^{-1}\left(\frac{15}{11}\right)\:=\:53.74616226 \:\approx\;\)54°

 

[oshea.emma]

wow,
How did you get so good at maths?

 

[jonboy]

wow,
How did you get so good at maths?

Well he post on MMB, this website, and Sparknotes consistently so he keeps in touch w/his math sustenance. Plus he is a retired math professor and is a math major. He has got it in him.

 

[oshea.emma]

fantastic!!

 

[jonboy]

fantastic!!

Yes and you can be fantastic at math as well if you put forth the commitment. I'd say that is fantastic as well!

 

[oshea.emma]

 

[jonboy]

Grrrrrrreat!