So I wanted to prove in an alternative way that the midpoint of two given points are collinear. The first way was as follows : Let $A,B,C\in \mathbb{R}^2$ be given by $A=(x_1,y_1)$, $B=(x_2,y_2)$ and $C=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$ : the line between A and B has slope $\frac{y_2-y_1}{x_2-x_1}$. The line between A and C has slope $\frac{\frac{y_1+y_2}{2}-y_1}{\frac{x_1+x_2}{2}-x_1}=\frac{y_2-y_1}{x_2-x_1}$ so it follows that they are collinear, because the slope is the same. Now I approached a different way in which I plugged $x=\frac{x_1+x_2}{2}$ in the line y = $\frac{y_2-y_1}{x_2-x_1}x$ (for simplicity the line goes through the origin). Then we get $y=\frac{y_2-y_1}{x_2-x_1}\frac{x_1+x_2}{2}$. I tried almost everything to arrive at $\frac{y_1+y_2}{2}$, but couldn't. Do you guys have any tips on how to tackle problems and in particular this one.
Proof that the midpoint of two given points are collinear
analytic geometryeuclidean-geometrygeometry
Best Answer
If one accepts that lines in the coordinate plane consist of all points $(x,y)$ which satisfy an equation of the form $ax + by = c$ for some constants $a,b,c$, then the proof goes like this:
Any two points $A, B$ determine a line: If $A = (x_A, y_A)$ and $B = (x_B, y_B)$, then we can set $a = (y_A - y_B), b = (x_B - x_A), c = y_Ax_B - y_Bx_A$. A bit of algebra will then show that both $A$ and $B$ satisfy $ax + by = c$.
If for some $a,b,c$, both $A$ and $B$ satisfy $ax + by = c$, we have $$ax_A + by_A = c\\ax_B + by_B = c.$$ Summing the two equations together and using the distributive rule to regroup gives $$a(x_A + x_B) + b(y_A + y_B) = 2c\\a\left(\frac{x_A + x_B}2\right) + b\left(\frac{y_A + y_B}2\right) = c.$$ So the midpoint $\left(\frac{x_A + x_B}2, \frac{y_A + y_B}2\right)$ also lies on the same line.