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?

$$d_1 = \sqrt{\left(\frac{x_1+x_2}{2}-x_1\right)^2 + \left(\frac{y_1+y_2}{2}-y_1\right)^2}
=\sqrt{\frac1{\color{red}4}\left(x_1-x_2\right)^2+ \frac1{\color{red}4}\left(y_1-y_2\right)^2}\\
=\frac12\sqrt{\left(x_1-x_2\right)^2+ \left(y_1-y_2\right)^2}.$$
