15 points! i have tried multiple times on this question and i don't understand it. matikas 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. enter the answers, in simplified form, in the boxes to complete the proof.

coordinate of point A is (a,b)

coordinate of point C is (2c,d)

slope of AD and BC =

Step-by-step explanation:

According to midpoint formula

If we have points P and Q

then the coordinate (x,y) of mid point of line PQ is given by

and

now from the given diagram A is the mid point of line joining  R(0,0) and S(2a,2b)

Using midpoint formula, coordinates of point A is given by

and

so we have

coordinate of point A is (a,b)

Also C is the midpoint of line joining T (2c,2d) and V(2c,0)

coordinate of point C is given by

and

so we have

coordinate of point C is (2c,d)

it is given that coordinate of point B is (a+c, b+d) and coordinate of D is (c,0)

If we have points P and Q

then the slope of PQ =

hence slope of AD=

                                 =

                                  =

                                  =

and slope of BC  =

                            =

                             =

                              =

so we have

slope of AD and BC =

Co-ordinates of A are ( a, b )

Co-ordinates of C are ( 2c, d )

Slope of line segments AD and BC is

Step-by-step explanation:

We know that, the co-ordinates of the mid-point of a line segment having end points (x,y) and is

Now as 'A' is the mid-point of the line segment RS having end points (0,0) and (2a,2b).

The co-ordinates of A will be ( , ) i.e

Now as 'C' is the mid-point of the line segment TV having end points (2c,2d) and (2c,0).

The co-ordinates of C will be ( , ) i.e. i.e. .

Further, we need to find the slope of line segments AD and BC.

AD has end points and . Then the slope of AD will be i.e

Similarly, BC has end points and . Then the slope of BC will be i.e

Hence, the slope of AD and BC is .

  5.   y=6; de=7; fg=8

  6.   x=4.2; jl=14.2; mn=13

step-by-step explanation:

when chords cross inside a circle, the product of the segment lengths is a constant.

5. dh×eh = fh×gh

  y = gh = dh×eh/fh = 3×4/2 . . . divide by the coefficient of gh

  y = 6

of course, the total chord length is the sum of the lengths of its segments.

  de = dh +eh = 3+4 = 7

  fg = fh +gh = 2+6 = 8

6. this problem works the same way as the previous one.

  x = lk = mk×nk/jk = 6×7/10 = 4.2

  jl = jk +lk = 10 +4.2 = 14.2

  mn = mk +nk = 6 +7 = 13

comment on chords and secants

you can think of the points h (problem 5) and k (problem 6) as points where the chords meet. a chord is part of a secant line, a line that intersects the circle in two points. one can also consider these points (h or k) to be the points where the respective secant lines meet.

then, the product that is a constant is the product of the distance from the meeting point (h or k) to one circle intersection with the distance from that meeting point to the other circle intersection. for problem 5, the constant is the product hd×he = hf×hg.

it turns out that this rule regarding the products of lengths to points of intersection is also true when the secants intersect outside the circle.

that is, you only need to remember one rule to work both kinds of problems.