proof

[red and white kop!]

two lines have equations y=m(1)x + c(1) and y = m(2)x + c(2), and m(1)m(2) = -1, proove that the lines are perpendicular.
im aware that this is supposed to be an extension question but still i dont know what to use in order to proove this, seeing that m(1)m(2) = -1 is apparently not enough. can somebody help with a clear proof on how to do this?

 

[mmm4444bot]

There are many different proofs (see Google).

HERE IS ONE

 

[red and white kop!]

thanks!

 

[mmm4444bot]

Here's another.

[attachment=0:w5aqia6b]junk4.JPG[/attachment:w5aqia6b]

 

[red and white kop!]

oh thats good! i had tried to proove it this way but c1 and c2 had stumped me; thanks again

 

[daon]

And another

Certainly they are not parallel, so they must intersect. WLOG assume m1 > m2.

We obtain the intersection point via:

$\displaystyle m_1x + c_1 = y = m_2x + c_2 \,\, \implies \,\, x = \frac{c_2-c_1}{m_1-m_2}$

$\displaystyle \,\, \implies \,\, y = m_1 \frac{c_2-c_1}{m_1-m_2} +c1 = m_2 \frac{c_2-c_1}{m_1-m_2} +c2$

This is not necessary, but we should know it exists.

We may form two vectors along the lines like so (x and y exist as the above quantities):

$\displaystyle P_1 = (x,y) = Q_1, P_2 = (x+1,y+m_1), Q_2 = (x+1,y+m_2)$

(You can verify the above by pluggin in (x+1) in for x and seeing it comes out to y+slope)

Notice $\displaystyle \bar{P}$ is a vector lying on the first line, $\displaystyle \bar{Q}$ on the second. To verify, you may take any point on the vector within the domain (x,x+1) and show it satisfies the linear equation.

$\displaystyle \bar{P} = <1, m_1>$
$\displaystyle \bar{Q} =<1, m_2>$

Taking the dot product we see:

$\displaystyle \bar{P} \cdot \bar{Q} = 1 + m_1m_2 = 1 + (-1) = 0$

Hene vectors $\displaystyle \bar{P}$ and $\displaystyle \bar{Q}$ are orthogonal, and so are the lines.