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Quadrilateral EFGH has coordinates E(a, a), F(3a, a), G(2a, 0), and H(0, 0). Find the midpoint of EF? Please show how the problem is done.

Respuesta :

angelicdragon12
You have to use the Midpoint formula:
[tex]M=( \frac{x_{2}-x{1}}{2}, \frac{y_{2}-y_{1}}{2}) [/tex]

We will use F as point 2 since is has the bigger x-value.
so then F=[tex](x_{2}, y_{2})[/tex]
and E=[tex](x_{1}, y_{1})[/tex]

[tex]M=( \frac{3a-a}{2}, \frac{a-a}{2}) [/tex]
[tex]M=( \frac{2a}{2} , \frac{0}{2})[/tex]
[tex]M=( a, 0)[/tex]

Your midpoint of EF is (a, 0)

Answer:

The midpoints of the EF is (2a,a) .

Step-by-step explanation:

Definition of midpoints.

The midpoint is the point lie in middle of a line segment. It is equidistant from both endpoints .

Formula

[tex]Midpoints = (\frac{x_{2}+x_{1})}{2} ,\frac{(y_{2}+y_{1}}{2})[/tex]

Quadrilateral EFGH has coordinates E(a, a), F(3a, a), G(2a, 0), and H(0, 0).

Now find out the midpoints of the  E(a, a) and F(3a, a) .

[tex]Midpoints\ of\ EF = (\frac{(3a+a)}{2} ,\frac{(a+a)}{2})[/tex]

[tex]Midpoints\ of\ EF = (\frac{4a}{2} ,\frac{2a}{2})[/tex]

Midpoints of EF = (2a,a)

Therefore the midpoints of the EF is (2a,a) .



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