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

Respuesta :

Panoyin
[tex]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)[/tex]
abidemiokin

The midpoint M of DE is at (1, 2) if DE has endpoints D(-1,6)and E(3,-2)

Midpoint of a line is the point that bisects or divides the line into two equal parts.

The formula for calculating the midpoint given the coordinate points D(-1,6)and E(3,-2) is expressed as:

[tex]M(X, Y) = (\frac{x_1+x_2}{2},\frac{y_1_y_2}{2} )[/tex]

X = -1+3/2

X = 2/2

X = 1

Similarly:

Y = 6+(-2)/2

Y = 6-2/2

Y = 4/2

Y = 2

Hence the midpoint M of DE is at (1, 2)

Learn more here:https://brainly.com/question/12223533

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