Question

15 Points!

Answer 1

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

Answer 2

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