If $(2, 1)$, $(3, 3)$ and $(6, 2)$ are the middlepoints of a sides of a triangle, what are the coordinates of a vertices of a triangle?
This part of the book deals with midpoints, with formula:
$$M_x = \frac{x_1 + \lambda \cdot x_2}{1 + \lambda}$$
So using this and equation of line length i should find vertices.
The solutions are: $(-1, 2)$, $(5, 0)$, $(7, 4)$
Hint: Write it out algebraically. If the vertices are of the form $(x_a, y_a)$, what are the vertices of the midpoints?
Now solve the system of equations.
If the vertices of the triangle are $(x_a, y_a), (x_b, y_b) $ and $(x_c, y_c)$, then we know that the midpoints are $ (-1, 2) = ( \frac{x_a + x_b} { 2}, \frac{ y_a + y_b } { 2} ), (5,0) = ( \frac{x_b + x_c} { 2}, \frac{ y_b + y_c } { 2} )$ and $(7,4) = ( \frac{x_c + x_a} { 2}, \frac{ y_c + y_a } { 2} )$
This allows us to solve for $x_a = -1 + 7 -5 = 1, x_b = 5 +(-1) - 7 = -3, x_c = 7+5 - (-1) = 13$ and $y_a = 2 + 4 - 0 = 6, y_b = 0 +2 - 4 = -2, y_c = 4 + 0 - 2 = 2 $.
Note, when you found the answer, interpret it geometrically and see if you can obtain a quick answer.
As a hint, the centroid of a triangle is the same as the centroid of the median triangle.
If $D, E, F$ are the midpoints, then the centroid is $\frac{D+E+F}{3}$, which tells us that the coordinates of $A$ must satisfy $ \frac{D+E+F}{3} = \frac{2}{3} D + \frac{1}{3} A$, so $A = E+F- D$.
While solving another question i got a solution.
I took a formula for Middlepoint of a line
x = 1/2(x1 + x2) and y = 1/2(y1 + y2)
and did something from long-time forgotten high school math,
x = 1/2(x1 + x2)
1/2(x1 + x2) = x
x1 + x2 = 2x
x1 = 2x - x2
and the same for y.
This yielded correct solutions.
All in all, it took me 3.5 hrs. After 2hrs i asked a question here. No luck this time.
$\displaystyle{\mbox{Let's define}\quad% \vec{M}_{1} \equiv \left(2,1\right)\,,\quad \vec{M}_{2} \equiv \left(3,3\right)\,,\quad\vec{M}_{3} \equiv \left(6,2\right). \vec{V}_{1} }$. The triangle vertices are given by $\vec{v}_{1}$, $\vec{v}_{2}$ and $-\vec{v}_{1} - \vec{v}_{2}$ which satisfy $$ \left({\tt I}\right)\quad {\vec{v}_{1} + \vec{v}_{2} \over 2} = \vec{M}_{1}\,, \qquad \left({\tt II}\right)\quad {\vec{v}_{1} + \vec{v}_{3} \over 2} = \vec{M}_{2}\,, \quad \left({\tt III}\right)\quad {\vec{v}_{2} + \vec{v}_{3} \over 2} = \vec{M}_{3} $$
Then, $\left({\tt II}\right) - \left({\tt III}\right)$ yields $\left(\vec{v}_{1} - \vec{v}_{2}\right)/2 = \vec{M}_{2} - M_{3}$ which we add to $\left({\tt I}\right)$. We get $\vec{v}_{1} = \vec{M}_{1} + \vec{M}_{2} - \vec{M}_{3} = \left(-1,2\right)$. From $\left({\tt I}\right)$, $\vec{v}_{2} = 2\vec{M}_{1} - v_{1} = \left(5,0\right)$. From $\left({\tt II}\right)$, $\vec{v}_{3} = 2\vec{M}_{2} - v_{1} = \left(7,4\right)$. In conclussion, $$\color{#ff0000}{\large% \vec{v}_{1} = \left(-1,2\right)\,, \qquad \vec{v}_{2} = \left(5,0\right)\,, \qquad \vec{v}_{3} = \left(-7,4\right)} $$
See the picture:
