Answer
${{P}_{1}}=\left( 3,-6 \right)$
Work Step by Step
Let
${{P}_{1}}=({{x}_{1}},{{y}_{1}})$and${{P}_{2}}=(7,-2)$
Here,
${{P}_{1}}=({{x}_{1}},{{y}_{1}})$ and ${{P}_{2}}=({{x}_{2}},{{y}_{2}})=(7,-2)$
The midpoint formula: The coordinate of the mid-point of two points ${{P}_{1}}=({{x}_{1}},{{y}_{1}})$ and${{P}_{2}}=({{x}_{2}},{{y}_{2}})$ is $M=(x,y)$, where $x=\frac{{{x}_{1}}+{{x}_{2}}}{2},\,y=\frac{{{y}_{1}}+{{y}_{2}}}{2}$.
Therefore, the coordinates of midpoint of the points ${{P}_{1}}=({{x}_{1}},{{y}_{1}})$ and ${{P}_{2}}=(7,-2)$ are $M=(x,y)$, where $x=\frac{{{x}_{1}}+7}{2},\,y=\frac{{{y}_{1}}-2}{2}$
The midpoint of the line segment joining the points ${{P}_{1}}$ and ${{P}_{2}}$ is $(5,-4)$.
Therefore, $x=5$ and $y=-4$.
$\Rightarrow 5=\frac{{{x}_{1}}+7}{2},\,\ and\,-4\,=\frac{{{y}_{1}}-2}{2}$
Solving each equation gives:
$5=\frac{x_1+7}{2}\\
10=x_1+7\\
x_1=3$
and
$4\,=\frac{y_1-2}{2}\\
-8=y_1-2\\
y_1=-6$
Thus,
${{P}_{1}}=\left( 3,-6 \right)$
