Question

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.

Answer 1

Answer:

coordinate of point A is (a,b)

coordinate of point C is (2c,d)

slope of AD and BC =$\frac{b}{a-c}$

Step-by-step explanation:

According to midpoint formula

If we have points P $(x_{1} ,y_{1} )$ and Q $(x_{2} ,y_{2} )$

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

$x=\frac{x_{1} +x_{2} }{2}$ and $y=\frac{y_{1} +y_{2} }{2}$

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

$x=\frac{0+2a}{2}= a$ and $y=\frac{0+2b}{2} =b$

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

$x= \frac{2c+2c}{2} =2c$ and $y=\frac{2d+0}{2} =d$

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 $(x_{1} ,y_{1} )$ and Q $(x_{2} ,y_{2} )$

then the slope of PQ =$\frac{y_{2} -y_{1} }{x_{2}-x_{1}  }$

hence slope of AD= $\frac{0-b }{c-a}$

                                 =$\frac{-b}{c-a}$

                                  =$\frac{-b}{-(a-c)}$

                                  =$\frac{b}{a-c}$

and slope of BC  =$\frac{d-(b+d)}{2c-(a+c)}$

                            =$\frac{d-b-d}{2c-a-c}$

                             =$\frac{-b}{c-a}$

                              =$\frac{b}{a-c}$

so we have

slope of AD and BC =$\frac{b}{a-c}$

Answer 2

Answer:

Co-ordinates of A are ( a, b )

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

Slope of line segments AD and BC is $\frac{-b}{c-a}$

Step-by-step explanation:

We know that, the co-ordinates of the mid-point of a line segment having end points (x,y) and $(x_{1} , y_{1} )$ is $(\frac{x+x_{1} }{2} , \frac{y+y_{1} }{2} )$

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 ( $\frac{0+2a}{2}$, $\frac{0+2a}{2}$ ) i.e $( a, b )$

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 ( $\frac{2c+2c}{2}$, $\frac{2d+0}{2}$ ) i.e. $( \frac{4c}{2} , \frac{2d}{2})$ i.e. $( 2c, d )$.

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

AD has end points $( a, b )$ and $( c, 0 )$. Then the slope of AD will be $\frac{0-b}{c-a}$ i.e $\frac{-b}{c-a}$

Similarly, BC has end points $( a+c, b+d )$ and $( 2c, d )$. Then the slope of BC will be $\frac{d-b-d}{2c-a-c}$ i.e $\frac{-b}{c-a}$

Hence, the slope of AD and BC is $\frac{-b}{c-a}$.