Area of region covered by the movement of a point This is a question from BDMO

In triangle ABC, AB= 12, BC=20, CA=16. X and Y
are two points in segment AB and AC
respectively. K is a point in segment XY , such
that XK/KY=7/5. If we let X and Y vary in
segment AB and AC , all the positions of K covers
a region. If we express the area of that region as
m/n in lowest term, compute m+n.

So, if we try to find the sides of the area, we should try putting the position of X and Y at the minimum and maximum possible point. But, I am quite out of idea how I can find the boundary of the area.
How can it be done?
 A: Let $A = (0,0)$ , B = (12, 0)$, C = (0, 16) $
Take point $X = (x, 0), Y = (0, y) $ where $0 \le x \le 12$ and $0 \le y \le 16$
Point $K$ is given by:
$K = \dfrac{5}{12} X + \dfrac{7}{12} Y $
Substituting $X$ and $Y$, then
$ K = \dfrac{1}{12} ( 5 x , 7 y ) $
The area of the region covered by $K$ is given by the integral
$ A = \displaystyle \int_{x = 0}^{12} \int_{y = 0}^{16} \| \dfrac{\partial K}{\partial x} \times \dfrac{\partial K}{\partial y} \| dy dx $
Now,
$\| \dfrac{\partial K}{\partial x} \times \dfrac{\partial K}{\partial y} \|= \dfrac{1}{144} \| (5, 0) \times (0, 7) \| = \dfrac{35}{144}$
Therefore, the area is given by
$ A = \bigg( \dfrac{35}{144} \bigg) \bigg( 12 \times 16 \bigg) = \dfrac{140}{3} $
A: A first remark is that the triangle is a multiple of the  famous $3-4-5$ right triangle.
As a consequence, we can work with orthogonal coordinates as indicated in the figure ($A(0,0), \ B(12,0), \ C(0,16)$). See the animated Geogebra figure here (use the horizontal and vertical sliders to move points $X$ and $Y$)

As $KX = 7 > KY = 5$,  $K$ is more attracted (coefficient 7) by $Y$ than by $X$ (coefficient 5), the resp. coefficients of attraction of $X$ and $Y$ are in the inverse order $5$ vs. $7$. Besides, these coefficients have to be normalized by their sum ($(\tfrac{5}{12},\tfrac{7}{12})$ instead of $(5,7)$ (a kind of percentage transformation), explaining that we can finally write :
$$K=\tfrac{5}{12}X+\tfrac{7}{12}Y=\tfrac{5}{12}\binom{12a}{0}+\tfrac{7}{12}\binom{0}{16b}\tag{1}$$
$$=\binom{5a}{28 b/3}=a\underbrace{\binom{5}{0}}_U+b\underbrace{\binom{0}{28/3}}_V\tag{2}$$
(for certain $a,b \in [0,1]$)
Relationship (2) shows that the locus is the rectangle (represented in red on the figure) generated by $U$ and $V$ above, with area :
$$7 \times \frac{5 \times 16}{12}=\frac{140}{3}$$
As it is an irreducible fraction, the answer is $143$.
