Given the points $A(4,4)$ and $B(7,0)$ which are vertices of the triangle $OAB$, where $O$ I assume is the origin, I have to find an equation of the line that bisects the angle $OBA$. There is a neat formula: $$\tan \alpha =\left|\frac{m_1-m_2}{1+m_1 m_2}\right|$$

Above is for acute angle between two straight lines. Where $\alpha$ is the angle between two straight lines and $m_1$ and $m_2$ are the slopes of those lines.

Since we have a point $B$, we only need a slope to form an equation of the line that bisects the angle. Now the slope can be written in two ways, thus forming an equation that is possible to be solved.

First equation: $$\left| \frac{0-m_2}{1}\right| = m_2$$ Second equation: $$\left| \frac{m_2+\frac{4}{3}}{1-\frac{4m_2}{3}}\right|$$

And then the slope should be found by equating the above two... But the answer is meaningless.. it is $i$.. And I have to keep it real.. Clearly.


Since $m_2$ is negative, you have to have $$\left|\frac{0-m_2}{1}\right|=\color{red}{-}m_2.$$

So, you'll have $$-m_2=\left|\frac{m_2+\frac 43}{1-\frac{4m_2}{3}}\right|$$ which should give $m_2=-\frac 12$.

  • $\begingroup$ But if I leave it negative then I will be looking at obtuse angle, no? And how is the modulus of negative $m_2$ equals $m_2$? $\endgroup$ – Naz Sep 14 '15 at 20:04
  • $\begingroup$ @isquared-KeepitReal: For the first, no, $m_2$ is the slope, not $\alpha$, so it can be negative. For the second, we have $|-m_2|=|m_2|$ and $|m_2|=-m_2$ because for $x\lt 0$, $|x|=-x$. $\endgroup$ – mathlove Sep 14 '15 at 20:09
  • $\begingroup$ It does help indeed. But how would I see myself, that the result of the first equation has to be negative. At the time of writing that equation I did not know whether $m_2$ would end up negative. I always have problems with modulus. $\endgroup$ – Naz Sep 14 '15 at 20:13
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    $\begingroup$ @isquared-KeepitReal: Setting $m_1=-\frac 43$ in $\left|(m_1-m_2)/(1+m_1m_2)\right|$ gives you $\left|\frac{-4/3-m_2}{1-(4m_2)/3}\right|$. So, you are right. Now note that we can have $|\frac{-4/3-m_2}{1-(4m_2)/3}|=|\frac{m_2+(4/3)}{1-(4m_2)/3}|$ because in general we have $|-A|=|A|$ for every $A\in\mathbb R$, which means that you can multiply the number inside the modulus by $-1$. $\endgroup$ – mathlove Sep 15 '15 at 10:56
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    $\begingroup$ Yes, that is what I was looking for. I see then, so there are indeed cases. And then if the answer to one case does not make sense it simply means it is not a valid solution. Perfect! This was actually just a part of the problem. Next you have to show that if C is the point at which bisector and OA meet, then it divides OA in the ratio OB:BA, which is equivalent to OC:CA. But I have solved that. I keep having issues with modulus, hopefully this has clarified everything. Thank you. $\endgroup$ – Naz Sep 15 '15 at 19:51

Here is an alternate method that should work:

Draw the line through $(4,4)$ perpendicular to the angle bisector, and let $(x,0)$ be the x-intercept of this line.

By congruent triangles, $\;\;d((x,0), (7,0))=d((4,4),(7,0))=5$, so $x=2$.

Then the midpoint $(3,2)$ of the line segment from $(4,4)$ to $(2,0)$ lies on the angle bisector,

so it has equation $y-0=\frac{2}{-4}(x-7)$ or $y=-\frac{1}{2}x+\frac{7}{2}$.

  • $\begingroup$ +1 for using geometric skill to solve an analytic-geometric problem. $\endgroup$ – Mick Sep 15 '15 at 2:32

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