# Trigonometry prove question

• May 23rd 2006, 01:35 AM
kingkaisai2
Trigonometry prove question
I should add this is desperate. The bracket after m and c just shows that the value of m and c are assumed to be different for each straight line.* means multiply.

Let a and b be the straight lines with equations y=m(1)x +c and y=m(2)x +c where m(1)*m(2) is not equal to 0. Use appropriate trigonometric formula to prove that a and b are perpendicular if and only if m(1)m(2) =1.

Please note that by appropriate- keep means to keep it as simple as possible- note this is high school maths so please try not to get too advance.
• May 23rd 2006, 01:16 PM
ThePerfectHacker
Quote:

Originally Posted by kingkaisai2
I should add this is desperate. The bracket after m and c just shows that the value of m and c are assumed to be different for each straight line.* means multiply.

Let a and b be the straight lines with equations y=m(1)x +c and y=m(2)x +c where m(1)*m(2) is not equal to 0. Use appropriate trigonometric formula to prove that a and b are perpendicular if and only if m(1)m(2) =1.

Please note that by appropriate- keep means to keep it as simple as possible- note this is high school maths so please try not to get too advance.

Actually you mean that,
$m_1\cdot m_2=-1$ :eek:
• May 23rd 2006, 02:24 PM
ThePerfectHacker
Let $\theta$ represent the angle a line forms with the x-axis. Now, this is important, there are two possible angles the one on the right and left. We let $\theta$ be the one on the right.

There is a theorem, given a line $y=mx+b$ then $\tan \theta =m$. This theorem is crucial in this proof.

Let us have two lines. One $y=m_1x+b_1$ crosses the x-axis at $M_1$ and the other $y=m_2x+b_2$ crosses the x-axis at, $M_2$. Let $P$ be the point of intersection of these two lines. Let us assume that they are perpendicular, meaning that $\not < M_1PM_2=90^o$. Let $\theta_1$ we the angle formed by the first line, and $\theta_2$ be the angle formed by the second line. Then, because we have a right triangle, $\theta_2=\theta_1+90^o$. Therefore, $\tan \theta_1 \tan (90^o+\theta_1)$ can expressed as, $-\tan \theta_1 \cot \theta _1$ cuz, $\tan (90^o+x)=-\cot x$ but because tangent and cotanget are reciprocals functions there product gives one thus,
$-\tan \theta_1 \cot \theta_1 =-1$
This means that,
$\tan \theta_1 \tan \theta_2=-1$
But, $m_1=\tan \theta_1, m_2=\tan \theta_2$
Thus,
$m_1m_2=-1$
• May 23rd 2006, 02:40 PM
ThePerfectHacker
For the converse,
--------
Let us have two lines. One $y=m_1x+b_1$ crosses the x-axis at $M_1$ and the other $y=m_2x+b_2$ crosses the x-axis at, $M_2$. Let $P$ be the point of intersection of these two lines. Let us assume that $m_1m_2=-1$. Let $\theta_1$ we the angle formed by the first line, and $\theta_2$ be the angle formed by the second line. Then, because we have a triangle, $\theta_2=\theta_1+P$. Where $P=\not < M_1PM_2$. Therefore, $m_1m_2=-1$ but, $m_1=\tan \theta_1$ and $m_2=\tan \theta_2$ thus,
$\tan \theta_1\tan (\theta_1+P)=-1$
Divide both sides by $\tan \theta_1$,
$\tan(\theta_1+P)=-\cot \theta_1$ because you get a reciprocal. Now using the same reason as above you can write, $\tan(\theta_1+P)=\tan(\theta_1+90^o)$
Thus,
$\theta_1+P=\theta_1+90^o$
Thus,
$P=90^o$.
Proof is complete
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Note (not for you): I know I could have not divided both sides sides by tangent without assuming the line was non-vertical and I cannot conclude,
$x=y$ from $\tan x=\tan y$ without denostrating this hold for angles $0\leq x\leq \pi$, i.e. domain of triangle.
So please do not bother me with these posts, I omitted that because that was high schools math.