Finding the Equations of Angle Bisectors The question is as follows: 

Let $K(5, 12)$, $L(14, 0)$ and  $M (0,0)$. The line $x + 2y = 14$ bisects angle $MLK$. Find equations for the bisectors of the angles $KML$ and $MKL$. 

Any help will be truly appreciated! 
 A: Let $KD$ be bisector of $\Delta KLM$.
Thus, since $$\frac{LD}{DM}=\frac{KL}{KM}=\frac{\sqrt{9^2+12^2}}{\sqrt{5^2+12^2}}=\frac{15}{13},$$
we obtain
$$D\left(\frac{13\cdot14+0}{15+13},0\right)$$ or
$$D(6.5,0).$$
Thus, $$m_{KD}=\frac{12-0}{5-6.5}=-8$$ and for the equation of $KD$ we obtain:
$$y-12=-8(x-5)$$ or
$$y=-8x+52.$$
Now, $y=-8x+52$ and $x+2y=14$ intersect in the point $I(6,4)$.
Thus, for the equation of the third bisector we obtain:
$$y-0=\frac{4-0}{6-0}(x-0)$$ or
$$y=\frac{2}{3}x.$$
Done!
A: First, let's find equations for line $MK$ which is easy: $y=\frac{12}{5}x$. Now the points that belong to bisector are equidistant from $MK$ and $ML$ (which is axis x). To find the slope we need to find tangent of one half of $\angle KML$. We can use trigonometric identity for that: $tan{\frac{x}{2}}=\sqrt{\frac{1-cosx}{1+cosx}}$. Thus, the equation for $\angle KML$ bisector is $y=\frac{2}{3}x$.  Now we need to find equation for line $KL$ which is $y=-\frac{4}{3}x+\frac{56}{3}$. The equation for bisector of $\angle KLM$ will be $$\large{0=\frac{\vert-\frac{4}{3}x-y+\frac{56}{3}\vert}{\sqrt{1+\frac{16}{9}}}}$$
A: Remember the parallelogram rule for adding a pair of vectors: their sum is the diagonal of the parallelogram with sides defined by the two vectors. The diagonal of a rhombus is also an angle bisector, so if we can make the two vectors the same length, then their sum will bisect the angle between them. One way to make this adjustment is to normalize the two vectors.  
For the given bisector, we have $\overrightarrow{LM}=M-L=(-14,0)$ and $\overrightarrow{LK}=K-L=(-9,12)$. Normalized, these become $(-1,0)$ and $\left(-\frac35,\frac45\right)$, respectively, and their sum is $\left(-\frac85,\frac45\right)$. This gives a slope of $\frac12$ for the bisector of $\angle{KLM}$. I assume that you can reconstruct the line with this information. For the other bisectors, you can do a similar computation. (For the third bisector, you could also use the fact that all three bisectors intersect at a common point.)  
I find it more convenient to work with the point-normal equation of a line for problems like these because it handles vertical lines without needing special cases for them. The diagonals of a rhombus are orthogonal, so we can get the normal to the angle bisector by subtacting one of the unit vectors from the other instead of adding them. For the given bisector, this would be $\left(-\frac35,\frac45\right)-(-1,0)=\left(\frac25,\frac45\right)$, so we can use $(1,2)$ as the normal of the line, producing the equation $(1,2)\cdot(x,y)=(1,2)\cdot(14,0)=14$, i.e., $x+2y=14$. 
