Question

PLEASE HELP ASAP!!!!

Hugo is writing a coordinate proof to show that the midpoints of a quadrilateral are the vertices of a parallelogram. He starts by assigning coordinates to the vertices of quadrilateral RSTV and labeling the midpoints of the sides of the quadrilateral as A, B, C, and D. Quadrilateral R S T V in a coordinate plane with vertex R at 0 comma 0, vertex S in the first quadrant at 2 a comma 2b, vertex T also in the first quadrant at 2 c comma 2 d, and vertex V on the positive side of the x-axis at 2 c comma 0. Point A is between points R and S, point B is between points S and T, point C is between points T and V, and point D is between points R and V. Enter the answers, in simplified form, in the boxes to complete the proof. The coordinates of point A are ( , ). The coordinates of point B are ( , ). The coordinates of point C are (2c, d) . The coordinates of point D are (c, 0) . The slope of both AB¯¯¯¯¯ and DC¯¯¯¯¯ is . The slope of both AD¯¯¯¯¯ and BC¯¯¯¯¯ is −bc−a . Because both pairs of opposite sides are parallel, quadrilateral ABCD is a parallelogram.

Answer 1

Explanation:

R(0,0)

A=((0+2a)/2,(0+2b)/2)=(a,b)

S(2a,2b)

B=((2a+2c)/2,(2b+2d)/2)=(a+c,b+d)

T(2c,2d)

C=((2c+2c)/2,(2d+0)/2)=(2c,d)

V(2c,0)

D=((2c+0)/2,(0+0)/2)=(c,0)

R(0,0)

slope of AB=(b+d-b)/(a+c-a)=d/c

slope of DC=(d-0)/(2c-c)=d/c

slope of AD=(0-b)/(c-a)=-b/(c-a)

slope of BC=(d-b-d)/(2c-a-c)=-b/(c-a)