Hmm, something you could use: $\displaystyle \tan\left(\dfrac{\pi}{2}-A\right) = \cot A$. So, you could use your formula to get the tangent of angle $\displaystyle \dfrac{\pi}{2}-A$ using $\displaystyle m_2$ and the slope of the hypotenuse.

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- Nov 16th 2013, 11:28 PMSlipEternalRe: Triangle problem in coordinate geometry
Hmm, something you could use: $\displaystyle \tan\left(\dfrac{\pi}{2}-A\right) = \cot A$. So, you could use your formula to get the tangent of angle $\displaystyle \dfrac{\pi}{2}-A$ using $\displaystyle m_2$ and the slope of the hypotenuse.

- Nov 17th 2013, 03:26 AMAaPaRe: Triangle problem in coordinate geometry
no! i did this right. A is the angle between the hypotenuse whose slope is 2/5 and m1.

then is used tan(90-A)= 1/tan A for m2 and and replaced m2 with -1/m2. to get two values of m1 with respect to tan A. - Nov 17th 2013, 09:02 AMSlipEternalRe: Triangle problem in coordinate geometry
Then, here are two possible reasons for the result:

1. You did not take into account the absolute value. Take the negative of one of those terms. Then you will probably get 0=0.

2. You are dividing by zero somewhere. Note that in the formula for tangent, if $\displaystyle m_1m_2 = -1$, the denominator is zero. Perhaps that can happen elsewhere in the problem, as well.

Edit: I just calculated it out. You did not take into account the absolute value. Calculating $\displaystyle \tan A$ with $\displaystyle m_1$ and $\displaystyle \dfrac{2}{5}$) or calculating $\displaystyle \dfrac{1}{\tan A}$ with $\displaystyle m_2 = -\dfrac{1}{m_1}$ and $\displaystyle \dfrac{2}{5}$ and solving for $\displaystyle m_1$ each time, I get the exact same result:

$\displaystyle m_1 = \dfrac{5\tan A \pm 2}{\pm 5 - 2\tan A}$ yielding the possible outcomes $\displaystyle m_1 = \dfrac{5\tan A + 2}{5-2\tan A}$ or $\displaystyle m_1 = \dfrac{2-5\tan A}{5+2\tan A}$. These cannot BOTH be true unless $\displaystyle \tan A = 0$. - Nov 17th 2013, 10:30 AMAaPaRe: Triangle problem in coordinate geometry
The absolute value in this formula is used to find the tanA of the acute angle as the acute angle can be formed by the lines in two ways. But as this is a right triangle. I dont think we need to take the modulus into account.

- Nov 17th 2013, 10:34 AMSlipEternalRe: Triangle problem in coordinate geometry
$\displaystyle \tan A = \pm \dfrac{\tfrac{2}{5} - m_1}{1+\tfrac{2}{5}m_1} = \pm \dfrac{2-5m_1}{5+2m_1}$

$\displaystyle 5\tan A + 2m_1\tan A = \pm (2-5m_1) = \pm 2 \mp 5m_1$

$\displaystyle 2m_1\tan A \pm 5m_1 = \pm 2 - 5 \tan A $

$\displaystyle (2\tan A \pm 5)m_1 = \pm 2-5\tan A$

$\displaystyle m_1 = \dfrac{\pm 2-5\tan A}{2\tan A \pm 5}$ - Nov 17th 2013, 10:44 AMAaPaRe: Triangle problem in coordinate geometry
The absolute value in this formula is used to find the tanA of the acute angle as the acute angle can be formed by the lines in two ways. But as this is a right triangle. I dont think we need to take the modulus into account.

- Nov 17th 2013, 10:58 AMSlipEternalRe: Triangle problem in coordinate geometry
If $\displaystyle m_1 > \dfrac{2}{5}$ or if $\displaystyle m_1<-\dfrac{5}{2}$, then $\displaystyle \dfrac{\tfrac{2}{5}-m_1}{1+\tfrac{2}{5}m_1}<0$. So, the plus/minus signs are needed. I don't understand what you mean by "take the modulus into account". What modulus?

- Nov 18th 2013, 07:35 AMAaPaRe: Triangle problem in coordinate geometry
m1 is making an acute angle A with the line of slope 2/5. Doesn't this mean tan A is positive?

- Nov 18th 2013, 09:39 AMSlipEternalRe: Triangle problem in coordinate geometry
Yes, so if $\displaystyle -\dfrac{5}{2} < m_1 < \dfrac{2}{5}$, you would use a plus sign for, otherwise, a minus sign. This way, $\displaystyle \tan A$ will be positive.

- Nov 18th 2013, 03:54 PMAaPaRe: Triangle problem in coordinate geometry
Yes but when it makes an acute angle doesn't it automatically mean that m1 lies between -5/2 and 2/5 ?

- Nov 18th 2013, 04:19 PMSlipEternalRe: Triangle problem in coordinate geometry
When two lines intersect, they form four angles. Unless they are all right angles, two of them will be obtuse and two will be acute. That is why the formula you are given takes the absolute value. It cannot be ignored.

- Nov 18th 2013, 05:37 PMSlipEternalRe: Triangle problem in coordinate geometry
Here is a better way to explain it. Suppose $\displaystyle m_1=-1$. That is a possible value where the formula is positive. Then $\displaystyle m_2=1$ which is not within the range that would allow you to use positive values. In fact, given any $\displaystyle -\dfrac{5}{2} < m_1 < \dfrac{2}{5}$, $\displaystyle m_2<-\dfrac{5}{2}$ or $\displaystyle \dfrac{2}{5} < m_2$

- Nov 19th 2013, 01:46 AMAaPaRe: Triangle problem in coordinate geometry
oh thank you! I got it. The two lines are making both acute and obtuse angles with each other. foolish me. thanks a lot for helping me out.