Answer
$P_1=(3,-6)$
Work Step by Step
The midpoint $M=(x,y)$ of the line segment from $P_{1}=(x_{1},y_{1})$ to $P_{2}=(x_{2},y_{2})$ is
$M=(x,y)=(\displaystyle \frac{x_{1}+x_{2}}{2},\frac{y_{1}+y_{2}}{2})$
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Given
$P_{1}=(x_{1},y_{1})$
$P_{2}=(x_{2},y_{2}),=(7,-2)$
$M=(\displaystyle \frac{x_{1}+x_{2}}{2},\frac{y_{1}+y_{2}}{2}) =(5,-4)$
$ \begin{aligned}
& & & & & & \\
\frac{x_{1}+7}{2} & =5 & /\times 2 & & \frac{y_{1}+(-2)}{2} & =-4 & /\times 2\\
x_{1}+7 & =10 & /-7 & & y_{1}-2 & =-8 & /+2\\
x_{1} & =3 & & & y_{1} & =-6 &
\end{aligned}$
$P_1=(3,-6)$
