so if you check the picture below, each side of the diamond has 90ft, so half-way from first to second is 45ft over the "x" line
a)
in this case, notice that "y" is static, is a constant, it doesn't change, whilst the "r" distance is and the "x" as well, when the runner is going to 1st base
[tex]\bf r^2=x^2+y^2\implies r^2=x^2+90^2\impliedby \textit{since "y" is a constant}
\\\\\\
2r\cfrac{dr}{dt}=2x\cfrac{dx}{dt}+0\implies \cfrac{dr}{dt}=\cfrac{x\frac{dx}{dt}}{r}\quad
\begin{cases}
x=45\\
\frac{dx}{dt}=19\\
r=45\sqrt{5}
\end{cases}
\\\\\\
\cfrac{dr}{dt}=\cfrac{19\cdot 45}{45\pm\sqrt{5}}\implies \cfrac{dr}{dt}=-\cfrac{19}{\sqrt{5}}\impliedby
\begin{array}{llll}
\textit{we used }-\sqrt{5}\\
\textit{because the rate is}\\
negative
\end{array}[/tex]
b)
well on this case, the distance "y" from home plate to 3rd base, isn't changing either, is a constant and is also 90ft as well
[tex]\bf r^2=x^2+y^2\implies r^2=x^2+90^2\impliedby \textit{since "y" is a constant}
\\\\\\
2r\cfrac{dr}{dt}=2x\cfrac{dx}{dt}+0\implies \cfrac{dr}{dt}=\cfrac{x\frac{dx}{dt}}{r}\quad
\begin{cases}
x=45\\
\frac{dx}{dt}=19\\
r=45\sqrt{5}
\end{cases}
\\\\\\
\cfrac{dr}{dt}=\cfrac{19\cdot 45}{45\pm\sqrt{5}}\implies \cfrac{dr}{dt}=\cfrac{19}{\sqrt{5}}\impliedby
\begin{array}{llll}
\textit{we used }+\sqrt{5}\\
\textit{because the rate is}\\
positive
\end{array}[/tex]
