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.
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.
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 $\displaystyle \tan A = 0$.
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.
$\displaystyle \tan A = \pm \dfrac{\tfrac{2}{5} - m_1}{1+\tfrac{2}{5}m_1} = \pm \dfrac{2-5m_1}{5+2m_1}$
$\displaystyle 5\tan A + 2m_1\tan A = \pm (2-5m_1) = \pm 2 \mp 5m_1$
$\displaystyle 2m_1\tan A \pm 5m_1 = \pm 2 - 5 \tan A $
$\displaystyle (2\tan A \pm 5)m_1 = \pm 2-5\tan A$
$\displaystyle m_1 = \dfrac{\pm 2-5\tan A}{2\tan A \pm 5}$
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.
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?
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.
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 $\displaystyle \dfrac{2}{5} < m_2$