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!