<s>[INST] How does $d_1$ equal $\frac{1}{2}\sqrt{\left(x_2-x_1\right)^2 + \left(y_2-y_1\right)^2}$? I'm going over the proof of the midpoint formula and the solution in my textbook solves its first distance as follows
$$d_1 = \sqrt{\left(\frac{x_1+x_2}{2}-x_1\right)^2 + \left(\frac{y_1+y_2}{2}-y_1\right)^2}$$
$$=\frac{1}{2}\sqrt{\left(x_2-x_1\right)^2 + \left(y_2-y_1\right)^2}$$
How does $d_1$ equal $\frac{1}{2}\sqrt{\left(x_2-x_1\right)^2 + \left(y_2-y_1\right)^2}$ ?
I tried to do it on paper but end up with $d_1$ = $\sqrt{\frac{1}{2}\left(x_2-x_1\right)^2 + \frac{1}{2}\left(y_2-y_1\right)^2} = \sqrt{\frac{1}{2}\left(\left(x_2-x_1\right)^2 + \left(y_2-y_1\right)^2\right)}$
I'm sure you cant just factor out a multiple right?
 [/INST] You forgot that the $\frac{1}{2}$ had to be squared when you were factoring it out:
$$
d_1 = \sqrt{\left(\frac{x_1+x_2}{2}-x_1\right)^2 + \left(\frac{y_1+y_2}{2}-y_1\right)^2}=
\sqrt{\left(\frac{x_1+x_2}{2}-\frac{2x_1}{2}\right)^2 + \left(\frac{y_1+y_2}{2}-\frac{2y_1}{2}\right)^2}=\\
\sqrt{\left(\frac{x_1+x_2-2x_1}{2}\right)^2 + \left(\frac{y_1+y_2-2y_1}{2}\right)^2}=\\
\sqrt{\frac{\left(x_2-x_1\right)^2}{2^2} + \frac{\left(y_2-y_1\right)^2}{2^2}}=
\sqrt{\frac{1}{4}\left(x_2-x_1\right)^2 + \frac{1}{4}\left(y_2-y_1\right)^2}=\\
\sqrt{\frac{1}{4}}\sqrt{\left(x_2-x_1\right)^2 + \left(y_2-y_1\right)^2}=
\frac{1}{2}\sqrt{\left(x_2-x_1\right)^2 + \left(y_2-y_1\right)^2}
$$
</s>