# Explaination behind the relation between the correlataion coefficient and distance of a typical point from the standard deviation line

I am reading a textbook on statists by Freedman, Pisani, and Purves. In one of the chapters about correlation between two variables, it is given that the vertical distance of a typical point from the standard deviation line (say $$s_v$$) on a scatter plot is, $$s_v=\sqrt{2(1-|r|)} \times \sigma_v$$, where $$r$$ is the correlation coefficient, and $$\sigma_v$$ is the vertical standard deviation.

How to work out this formula mathematically? I thought about working it out from the slope of the SD line, or formula for $$r$$ but I can't seem to work it out or find a hint/solution online. It's not a homework problem it's a sort of technical footnote.

It is also mentioned that there are similar formulas for the horizontal direction. Is $$s_h=\sqrt{2(1-r)} \times \sigma_h$$, the formula for the horizontal distance? If not, what will it be?

Give $$(x_i,y_i)_{i=1}^n$$, let $$d_i$$ be the vertical distance from SD line to $$(x_i,y_i)$$. Then we have $$d_i = y_i - \left(\frac{\sigma_y}{\sigma_x}(x_i-\bar{x})+\bar{y}\right).$$
Thus r.m.s(vertical distance of all data) is given by \begin{align} \sqrt{\frac{1}{n}\sum d_i^2}=SD[d_i]&=SD[y_i - \left(\frac{\sigma_y}{\sigma_x}(x_i-\bar{x})+\bar{y}\right)]\\ &=SD[y_i - \frac{\sigma_y}{\sigma_x}x_i + const]. \end{align} SD is invariant with constant shift, applying $$SD(X+Y)=\sqrt{Var(X)+Var(Y)+2COV(X,Y)}$$ onto the above: \begin{align} SD[y_i - \frac{\sigma_y}{\sigma_x}x_i]&=\sqrt{\sigma_y^2+\frac{\sigma_y^2}{\sigma_x^2}\sigma_x^2 -2\frac{\sigma_y}{\sigma_x}COV(x,y)}\\ &=\sqrt{2\sigma_y^2 -2\frac{\sigma_y}{\sigma_x}r\sigma_y\sigma_x}\\ &=\sqrt{2(1-r)}\sigma_y \end{align}