The problem goes as follows: Given a line $l:y=ax$, where $a\in \Bbb{R}$ and $a>0$, and a point $P(b,c)$ that does not lie on $l$. Let $M$ be a point on the x-axis, and $Q$ the intersection of line $l$ with line $PM$.
If $O$ denotes the origin, what are the coordinates of point $M$ such that triangle $OMQ$ has minimal area?
Now, I've been able to solve some particular cases for specific points, and the conjecture I got is that when $M$ is located in such a way that $|PM|=|PQ|$, the solution is obtained. I'm not sure how to obtain the a geometric or analytical justification (it seems very complex trying to obtain a function to maximize). I'd appreciate some help! thanks in advance.
Edit: When $M$ is at the right or left of the point $(0, c-ab)$: the area of the triangle can be made arbitrarily close to $0$ or the area of the triangle will reach a local minimum. The interesting part of the problem is to find that local minimum.
So we have point $P(b,c)$ and $M(m,0)$. The equation of the line between them is $$\frac{x-b}{m-b}=\frac{y-c}{0-c}$$ The intersection of this line with $l$ is $$y=ax$$ When we put these together we get $$\frac{x-b}{m-b}=\frac{ax-c}{0-c}$$ or $$-cx+bc=amx-cm-abx+bc$$ Extracting $x$ (the x coordinate of the $Q$ point): $$x_Q=\frac{cm}{am-ab+c}$$ The corresponding $y$ coordinate is $$y_Q=\frac{acm}{am-ab+c}$$ The area of the triangle $OMQ$ is $$A=\frac 12\frac{am^2c}{am-ab+c}$$ since $y_Q$ is the height and $m$ is the base. Take the derivative with respect to $m$ and equate it to $0$. One solution is $m=0$. The other is $$m=\frac{2}{a}(ab-c)$$ Now just plug in this $m$ into the equations for $x_Q$, $y_Q$, then calculate $|PM|$ and $|PQ|$ and you get the answer. If you calculate $PM$ and $PQ$ as vectors, you should get $PM=-PQ$.