How does $d_1$ equal $\frac{1}{2}\sqrt{\left(x_2-x_1\right)^2 + \left(y_2-y_1\right)^2}$?

Question

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?

Answer 1

You forgot that the $\frac{1}{2}$ had to be squared when you were factoring it out:

$$ d_1 = \sqrt{\left(\frac{x_1+x_2}{2}-x_1\right)^2 + \left(\frac{y_1+y_2}{2}-y_1\right)^2}= \sqrt{\left(\frac{x_1+x_2}{2}-\frac{2x_1}{2}\right)^2 + \left(\frac{y_1+y_2}{2}-\frac{2y_1}{2}\right)^2}=\\ \sqrt{\left(\frac{x_1+x_2-2x_1}{2}\right)^2 + \left(\frac{y_1+y_2-2y_1}{2}\right)^2}=\\ \sqrt{\frac{\left(x_2-x_1\right)^2}{2^2} + \frac{\left(y_2-y_1\right)^2}{2^2}}= \sqrt{\frac{1}{4}\left(x_2-x_1\right)^2 + \frac{1}{4}\left(y_2-y_1\right)^2}=\\ \sqrt{\frac{1}{4}}\sqrt{\left(x_2-x_1\right)^2 + \left(y_2-y_1\right)^2}= \frac{1}{2}\sqrt{\left(x_2-x_1\right)^2 + \left(y_2-y_1\right)^2} $$

Answer 2

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