If de has endpoints D(-1,6)and E(3,-2),then the midpoint M of DE lies in Quadrant1

1 answer:

7 0

$M(\frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2}) \\M(\frac{-1 + 3}{2}, \frac{6 + (-2)}{2}) \\ M(\frac{2}{2}, \frac{6 - 2}{2}) \\ M(1, \frac{4}{2}) \\ M(1, 2)$

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