If the line 2x+3y = 4 is relected over the line 5x+6y=7 ....

[leilsilver]

The question is:

If the line 2x+3y= 4 is relected over the line 5x+6y=7, it becomes the line ax+by=c, where a and b are integers and teh greatest common factor of a and b is 1. Find a+b.

And I tried the trig identity of tan(a+b) and tan(a-b) from how my teacher told us to do it only doing it where the points intersect and imagine it as the origin and fining the angle betweent the initial line and the line to that is being reflected upon. The numbers were scrambled...And I tried finding the intersecting points and did what my dad did: he told me to subtract one point from the other i.e. the line 2x+3y= 4 minus the line 5x+6y=7. didn't work. I got a line twice as big from 2x+3y=4.

If anyone could help it would be greatly appreciated.

 

[soroban]

Re: If the line 2x+3y = 4 is relected over the line 5x+6y=7

Hello, leilsilver!

It's been many years since I've solved this particular problem, so I improvised a method.
It's similar to your game plan and your teacher's approach.

If the line \(\displaystyle 2x\,+\,3y\:=\:4\) is relected over the line \(\displaystyle 5x\,+\,6y\:=\:7\), it becomes the line \(\displaystyle ax\,+\,by\:=\:c\),

where \(\displaystyle a\) and \(\displaystyle b\) are integers and \(\displaystyle GCD(a,b)\,=\,1.\;\) Find \(\displaystyle a\,+\,b\)

The two lines intersect at \(\displaystyle (-7,6)\)

My diagram is upside-down from the actual graph of the lines,
\(\displaystyle \;\;\)but it gives the relative positions of the three lines.

      :            m
      :           *
      :                    * m2 = -5/6
      :        *        *
      :              *
      :     * θ2  *           *  m1 = -2/3
      :        *  θ1    *
      :  *  *     *
      :  *  *
      * - - - - - - - - - - - - - - - -
   (-7,6)

The two given lines have slope; \(\displaystyle \,m_1\,=\,-\frac{2}{3},\;m_2\,=\,-\frac{5}{6}\)

\(\displaystyle \;\;\\)Between them is angle \(\displaystyle \theta_1.\)


We want a third line with slope \(\displaystyle m\)

\(\displaystyle \;\;\)so that angle between \(\displaystyle m\) and \(\displaystyle m_2\) is \(\displaystyle \,\theta_2\,=\,\theta_1.\)


Using the difference-of-tangents formula you suggested,

\(\displaystyle \;\;\)we have: \(\displaystyle \,\tan\theta_1\:=\:\frac{m_2\,-\,m_1}{1\,+\,m_1\cdot m_2}\;=\;\frac{\left(-\frac{5}{6}\right)\,-\,\left(-\frac{2}{3}\right)}{1\,+\,\left(-\frac{5}{6}\right)\left(-\frac{2}{3}\right)} \;= \;
-\frac{3}{28}\)

Then: \(\displaystyle \,\tan\theta_2\:=\:\frac{m\,-\,\left(-\frac{5}{6}\right)}{1\,+\,(m)\left(-\frac{5}{6}\right)} \;=\;-\frac{3}{28}\)

Multiply top and bottom by 6: \(\displaystyle \:\frac{6m\,+\,5}{6\,-\,5m}\:=\:-\frac{3}{28}\)

\(\displaystyle \;\;\)and we get: \(\displaystyle \:168m\,+\,140\:=\:-18\,+\,15m\;\;\Rightarrow\;\;153m\,=\,-158\;\;\Rightarrow\;\;m\,=\,-\frac{158}{153}\)


The line contains \(\displaystyle (-7,6)\) and has slope \(\displaystyle m\,=\,-\frac{158}{153}\)

Its equation is: \(\displaystyle \,y\,-\,6\;=\;-\frac{158}{153}(x\,+\,7)\;\;\Rightarrow\;\;153y\,-\,918\;=\;-158x\,-\,1106\)


Therefore: \(\displaystyle \:158x\,+\,153y\:=\:-188\;\;\Rightarrow\;\;a\,+\,b\;=\;158\,+\,153\;=\;311\)


But someone check my reasoning and work . . . please!

 

[pka]

Re: If the line 2x+3y = 4 is relected over the line 5x+6y=7

But someone check my reasoning and work . . . please!

The lines insect at (-1,2) not (-7,6)

 

[leilsilver]

Yeah...the line does intersect at (-1,2) not (-7,6) ....But I'll try your reasoning with the correct numbers...And we'll see what I get. And if you really just reflect it without the use of ...math...it's actually y=-x +1. ...I just can't do it with a proper equation! Which bugs the **** out of me...because my math exam is like...on wednesday...