Question

parallelogram abcd has coordinates of A (7,1), B (-2,-3), and C (0,3). What must be the coordinates of point d? Explain how you found your answer.

Answer 1

Answer:(9,7)

Step-by-step explanation:

(A-B)=(7--2,1--3)=(9,4), displacement vector B to A

(A-B) + B = (9,4)+(-2,-3)=(7,1) ✔

(A-B) + C = (9,4)+(0,3)=(9,7) edge CD || AB

(C-B) = (2,6) displacement vector B to C

(C-B) + A = (2,6)+(7,1)=(9,7) ✔ edge AD || CB

Answer 2

Answer: The coordinates of point D is (9,7).

Step-by-step explanation:

Given, A parallelogram ABCD has coordinates A (7,1), B (-2,-3), and C (0,3). .

To find : Coordinates of D.

Let coordinates of D be (x,y).

Since , Diagonals of a parallelogram bisects each other.

So, Mid point of AC = Mid point of BD { Both AC and BD are diagonals]

$\Rightarrow(\dfrac{7+0}{2},\dfrac{1+3}{2})=(\dfrac{-2+x}{2},\dfrac{-3+y}{2})\ \ [\text{Using Mid point formula}]\\\\\Rightarrow\ 7=-2+x\ \ \&\ \ 4=-3+y\\\\\Rightarrow\ x=7+2=9\ \ \&\ \ y=4+3=7$

Hence, the coordinates of point D is (9,7).