Given the following: $$L:x+y=1$$ $$M:3y=2x+6$$ $$N:2x+y=6$$ Find point $u,v,w\in\Bbb R^2$ such that in the triangle $[u,v,w]$, $L$ is the perpendicular bisector of $[u,v]$ (condition *), $M$ is the median from $u$ (condition $**$), and $N$ is the altitude from $v$ (condition ***).
Here's my approach:
$$L:x+y=1\implies y=1-x; \text{ so let } x=t, y=1-t$$ $$M:x=t, y=\frac{2t+6}{3}=\frac{2}{3}t+2$$ $$N:x=t, y=6-2t$$
In my understanding, I need to come up with six equations in order to find the coordinates $(u_1, u_2), (v_1, v_2), \text{ and } (w_1, w_2)$. I have come up with five equations:
$$\text{ 1. } (u-v)\cdot (t,1-t)=0 \text{ (condition *) }$$ $$\text{ 2. } \frac{1}{2}(u-v)= (t,1-t) \text{ (condition **) }$$ $$\text{ 3. } (w-u)\cdot (t, 6-2t)= 0 \text{ (condition ***) }$$ $$\text{ 4. } \|(w-u)\times (t, 6-2t)\|=\|(v-u)\times (t, \frac{2}{3}t+2)\| \text{ (median $M$ divides $[u,v,w]$ into two triangles with equal area) }$$ $$\text{ 5. } (t,6-2t)\cdot (u-w)=0$$
I would appreciate if someone could please give me the following insights:
- Is my approach correct?
- What should the 6th equation be?
- Is there a less tedious approach to finding these points?
- Another thing is that there should probably be more than six equations, since there are some variables like $s,t$, etc.
Your approach is obviously wrong, since five equations do not determine six variables. Instead, note that
Therefore, let $v=(k,6-2k)$. Reflecting this in $L$ makes $u=(2k-5,1-k)$, but $u$ must lie on $M$ so $$3(1-k)=2(2k-5)+6$$ $$3-3k=4k-10+6$$ $$7k=7\qquad k=1$$ So $u=(-3,0)$ and $v=(1,4)$. The perpendicular from $u$ to $N$ is $2y=x+3$; $w$ lies on this line, so let $w=(2q-3,q)$. The final constraint, that of $M$ being the median from $u$, is equivalent to the midpoint of $vw$ – $\left(q-1,\frac12q+2\right)$ – lying on $M$: $$3\left(\frac12q+2\right)=2(q-1)+6$$ $$\frac32q+6=2q-2+6$$ $$3q+12=4q+8\qquad q=4$$ In summary, $u=(-3,0)$, $v=(1,4)$ and $w=(5,4)$.