Question

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.

Answer 1

You have to use the Midpoint formula:

`M=( \frac{x_(2)-x{1}}{2}, (y_(2)-y_(1))/(2))`

We will use F as point 2 since is has the bigger x-value.
so then F=
`(x_(2), y_(2))`
and E=
`(x_(1), y_(1))`


`M=( (3a-a)/(2), (a-a)/(2))`

`M=( (2a)/(2) , (0)/(2))`

`M=( a, 0)`

Your midpoint of EF is (a, 0)

Answer 2

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

Explanation:

Definition of midpoints.

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

Formula

`Midpoints = ((x_(2)+x_(1)))/(2) ,((y_(2)+y_(1))/(2))`

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) .

`Midpoints\ of\ EF = (((3a+a))/(2) ,((a+a))/(2))`

`Midpoints\ of\ EF = ((4a)/(2) ,(2a)/(2))`

Midpoints of EF = (2a,a)

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