1. ## 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.

2. 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,
$\displaystyle m_1\cdot m_2=-1$

3. Let $\displaystyle \theta$ 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 $\displaystyle \theta$ be the one on the right.

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

Let us have two lines. One $\displaystyle y=m_1x+b_1$ crosses the x-axis at $\displaystyle M_1$ and the other $\displaystyle y=m_2x+b_2$ crosses the x-axis at, $\displaystyle M_2$. Let $\displaystyle P$ be the point of intersection of these two lines. Let us assume that they are perpendicular, meaning that $\displaystyle \not < M_1PM_2=90^o$. Let $\displaystyle \theta_1$ we the angle formed by the first line, and $\displaystyle \theta_2$ be the angle formed by the second line. Then, because we have a right triangle, $\displaystyle \theta_2=\theta_1+90^o$. Therefore, $\displaystyle \tan \theta_1 \tan (90^o+\theta_1)$ can expressed as, $\displaystyle -\tan \theta_1 \cot \theta _1$ cuz, $\displaystyle \tan (90^o+x)=-\cot x$ but because tangent and cotanget are reciprocals functions there product gives one thus,
$\displaystyle -\tan \theta_1 \cot \theta_1 =-1$
This means that,
$\displaystyle \tan \theta_1 \tan \theta_2=-1$
But, $\displaystyle m_1=\tan \theta_1, m_2=\tan \theta_2$
Thus,
$\displaystyle m_1m_2=-1$

4. For the converse,
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Let us have two lines. One $\displaystyle y=m_1x+b_1$ crosses the x-axis at $\displaystyle M_1$ and the other $\displaystyle y=m_2x+b_2$ crosses the x-axis at, $\displaystyle M_2$. Let $\displaystyle P$ be the point of intersection of these two lines. Let us assume that $\displaystyle m_1m_2=-1$. Let $\displaystyle \theta_1$ we the angle formed by the first line, and $\displaystyle \theta_2$ be the angle formed by the second line. Then, because we have a triangle, $\displaystyle \theta_2=\theta_1+P$. Where $\displaystyle P=\not < M_1PM_2$. Therefore, $\displaystyle m_1m_2=-1$ but, $\displaystyle m_1=\tan \theta_1$ and $\displaystyle m_2=\tan \theta_2$ thus,
$\displaystyle \tan \theta_1\tan (\theta_1+P)=-1$
Divide both sides by $\displaystyle \tan \theta_1$,
$\displaystyle \tan(\theta_1+P)=-\cot \theta_1$ because you get a reciprocal. Now using the same reason as above you can write, $\displaystyle \tan(\theta_1+P)=\tan(\theta_1+90^o)$
Thus,
$\displaystyle \theta_1+P=\theta_1+90^o$
Thus,
$\displaystyle P=90^o$.
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,
$\displaystyle x=y$ from $\displaystyle \tan x=\tan y$ without denostrating this hold for angles $\displaystyle 0\leq x\leq \pi$, i.e. domain of triangle.
So please do not bother me with these posts, I omitted that because that was high schools math.