Answer
A)The desired intersection point is $\left( 2.65,\,1.6 \right)$.
B)The distance from the point $\left( 1.4,1.3 \right)$ to the midpoint of the shortest side $\left( 2.65,\,1.6 \right)$ is $1.2855$.
Work Step by Step
(a)
The desired intersection point is the midpoint of the shortest side. End points of the shortest side of the error triangle are $\left( 2.7,\,1.7 \right)$ and $\left( 2.6,\,1.5 \right)$.
Here, ${{P}_{1}}=\left( {{x}_{1}},{{y}_{1}} \right)=\left( 2.7,\,1.7 \right)$ and ${{P}_{2}}=\left( {{x}_{2}},{{y}_{2}} \right)=\left( 2.6,\,1.5 \right)$
By the midpoint formula
$\left( x,\,y \right)=\left( \frac{2.7+2.6}{2},\,\frac{1.7+1.5}{2} \right)$
$\left( x,\,y \right)=\left( \frac{5.3}{2},\,\frac{3.2}{2} \right)$
$\left( x,\,y \right)=\left( 2.65,\,1.6 \right)$
Thus, the midpoint of the shortest side is $\left( 2.65,\,1.6 \right)$
Therefore, the desired intersection point is $\left( 2.65,\,1.6 \right)$
(b)
From part (a),
The midpoint of the shortest side is $\left( 2.65,\,1.6 \right)$.
The opposite vertex of the triangle is $\left( 1.4,1.3 \right)$.
To find the distance from the point $\left( 1.4,1.3 \right)$ to point $\left( 2.65,\,1.6 \right)$:
By using the distance formula
$d=\sqrt{{{\left( 2.65-1.4 \right)}^{2}}+{{\left( 1.6-1.3 \right)}^{2}}}$
$d=\sqrt{{{\left( 1.25 \right)}^{2}}+{{\left( 0.3 \right)}^{2}}}$
$d=\sqrt{1.5625+0.09}$
$d=\sqrt{1.6525}$
$d\approx 1.2855$
Therefore, the distance between the right fielder and second base is approximately $1.2855$
