In the coordinate plane, quadrilateral ABCD has vertices A(-2, 3), B(4,5), C(10, -1), and D(8, -9). Let E, F, G, and H be the midpoints of AB, BC, CD, and DA, respectively.
Prove that quadriateral EFGH is a parallelogram.
Enter your proof in the space provided.

 G(3, 1), H(2, 3)

Step-by-step explanation:

When D is the midpoint of EG, it means ...

 D = (E + G)/2

or

 G = 2D -E = 2(3,4) -(3,7) = (2·3-3, 2·4-7) = (3, 1)

Likewise, H is ...

 H = 2D -F = 2(3,4) -(4,5) = (2·3-4, 2·4-5) = (2, 3)

not my answer but i hope it helps you.

step-by-step explanation:

i think something is missing in your question because y-1=(x-12) would not pass through the point (-4,-3). because if you replace x by -4 your are supposed to find y=-3 which is not the case her. if you replace x by -4 you find =-15.

i assume the right equation is y-1=1/4*(x-12) because this one is passing through both your points (if you are interested to know how i found that out i can tell you).

in this equation if i replace x by -4 i find y=-3 and if i replace x by 12, i find y=1 so the line representing this equation is passing through those 2 points.

then the standard form of the equation is finding an expression of y in function of x

y-1=1/4*(x-12)

y=x/4-3+1

y=x/4-2

hope that .

do you confirm there was a mistake in the question?