Consider this small and simple dataset:

points $=$ pd$.$DataFrame$($

$[[1,1],[2,1],[3,3],[6,5],[7,7],[7,6]],$

columns$=["$X$","$y$"]$

$)$

base $=$ alt.Chart$($points$).$mark$\_$point$().$encode$($

x$=\text{'}$X$\text{'},$

y$=\text{'}$y$\text{'}$

$)$

base

Now consider these three $**$potential$**$ lines we could fit for the same dataset:

lines $=$ pd$.$DataFrame$($

$[$

$[0\mathrm{.}93,0\mathrm{.}017562,$ 'Line A$\text{'}],$

$[7,5\mathrm{.}9868,$ 'Line A$\text{'}],$

$[0,0\mathrm{.}1022,$ 'Line B$\text{'}],$

$[7,6\mathrm{.}965,$ 'Line B$\text{'}],$

$[0\mathrm{.}26,0\mathrm{.}003564,$ 'Line C$\text{'}],$

$[8,7\mathrm{.}0965,$ 'Line C$\text{'}]$

$],$

columns$=["$X$","$y$",$ 'Name'$]$

$)$

base $+$ alt.Chart$($lines$).$mark$\_$line$().$encode$($

x$=\text{'}$X$\text{'},$

y$=\text{'}$y$\text{'},$

color$=$'Name'

$)$

Use the graph below to roughly calculate the average squared vertical distance between the points and the blue line $("$Line A$"$ above$).**$Read values of the graph to a precision of $0\mathrm{.}25**($e$.$g$.1,1\mathrm{.}25,1\mathrm{.}5,1\mathrm{.}75,2).$ We reprint the plot for you with only a single line to make it easier to estimate the locations on the graph.

$*$Save your answer to a variable named $`$answer$2\_0`.*$

base $+$ alt.Chart$($lines$[$lines$[\text{'}$Name$\text{'}]==$ 'Line A$\text{'}]).$mark$\_$line$().$encode$($

x$=\text{'}$X$\text{'},$

y$=\text{'}$y$\text{'},$

$)$