Question

Becky is writing a coordinate proof to show that the diagonals of a parallelogram bisect each other. She starts by assigning coordinates as given .

Answer 1

Answer:

Proved.

Step-by-step explanation:

Given that coordinates of A, B and D are (0,0) (a,0) and (b,c) respectively

C is having y coordinate same as D but x coordinate will be a+b

(a+b,c) is C

Now the mid point of AC is using mid point formula

$(\frac{a+b}{2} ,\frac{c}{2} )$

to prove that the diagonals bisect each other it is sufficient to show that

The coordinates of the midpoint of diagonal BD=$(\frac{a+b}{2} ,\frac{c}{2} )$

Since we have B = (1,0) and D = (b,c)

using mid point formula we get mid point of BD

= $(\frac{a+b}{2} ,\frac{c}{2} )$

Comparing these two we find the mid point is the same for both the diagonals.

In other words, in a parallelogram diagonals bisect each other,

Answer 2

Answer:

Ok! So I've done this test, and i have the answers. They should all be correct, but let me know if you missed one so others can know!

Step-by-step explanation:

*The coordinates of point C are (a+b, c).

*The coordinates of the midpoint of diagonal AC¯¯¯¯¯ are ​ (a+b over 2, c over 2) ​ .

*The coordinates of the midpoint of diagonal BD¯¯¯¯¯ are (a+b over 2, c over 2).

*AC¯¯¯¯¯ and BD¯¯¯¯¯ intersect at point E with coordinates ( a+b over 2, c over 2)..

*By the definition of midpoint, AE¯¯¯¯¯≅ CE¯¯¯¯¯ and AE¯¯¯¯¯ ≅DE¯¯¯¯¯ .

*Therefore, diagonals AC¯¯¯¯¯ and BD¯¯¯¯¯ bisect each other.

This should be the answers for the Becky and Julia question! Again, make sure to let me know if anythings wrong. Also, if you want to give brainliest, I definitely wouldn't be mad.