Find $m>1/2$ that minimizes the area of the triangle formed by lines $y=10-2x$, $y=mx$, $y=-x/m$ I am tackling this problem below:

A triangle is formed by the three lines 
  $$\begin{align}
y &=10-2x \\ 
y &= mx \\
y   &=-\frac{x}{m}
\end{align}$$ where $m>\frac{1}{2}$. Find the value of $m$ for which the area of the triangle is a minimum.

My thoughts are these:
I tried to draw the three graphs nothe rectangular coordinate, and since $m> \frac{1}{2}$, $-\frac{1}{m}>2$, but after that how can I decide the area of triangle formed?      
 A: Let $\ell_1$ be the line with equation $y = 10-2x$, let $\ell_2$ be the line with equation $y = mx$, and let $\ell_3$ be the line with equation $y = \frac{-x}{m}$.  Then the lines $\ell_1$ and $\ell_2$ intersect at 
$$P = \left( \frac{10}{m+2}, \frac{10m}{m+2} \right),$$
and the lines the lines $\ell_1$ and $\ell_3$ intersect at
$$
Q = \left( \frac{-10m}{1-2m}, \frac{10}{1-2m} \right).
$$
If $O = (0,0)$, then the line segments $\overline{OP}$ and $\overline{OQ}$ are orthogonal, which means that the area of the triangle is given by
$$
A = \frac{1}{2} \left|OP\right| \left| OQ\right| 
= \frac{1}{2} \left( \frac{10 \sqrt{1+m^2}}{m+2} \right) \left( \frac{10 \sqrt{1+m^2}}{1-2m} \right) = \frac{50(1+m^2)}{(m+2)(1-2m)}.
$$
A: Although an answer is given, I want to add my answer too, which is less tedious. 
It is clear that the lines $y=mx$ and $y=\frac{-x}{m}$ are perpendicular and their point of intersection is the origin. 
The other side is $2x+y-10=0$
Since we need to calculate the area of the triangle, we can always use $\frac{h×b}{2}$ instead of coordinates. Perpendicular distance on the line from origin is $\frac{10}{\sqrt{5}}=2\sqrt{5}$.
Now we use that our triangle is right triangle. It is trivial from here. The minimum area is $(2\sqrt{5})^2=20$. 
