Answer with explanation:
Property of Quadrilateral A B CD
⇒Equation of Line AD is, y= -3 x.
⇒Equation of Line BC is, y= -3 x +11
Coordinates of CD = C(2,5) and D(-1,3)
Equation of line CD will be
![\frac{y-5}{x-2}=\frac{3-5}{-1-2}\\\\-3 \times (y-5)=-2\times (x-2)\\\\-3 y +15=-2 x +4\\\\2 x -3 y +11=0](https://tex.z-dn.net/?f=%5Cfrac%7By-5%7D%7Bx-2%7D%3D%5Cfrac%7B3-5%7D%7B-1-2%7D%5C%5C%5C%5C-3%20%5Ctimes%20%28y-5%29%3D-2%5Ctimes%20%28x-2%29%5C%5C%5C%5C-3%20y%20%2B15%3D-2%20x%20%2B4%5C%5C%5C%5C2%20x%20-3%20y%20%2B11%3D0)
Equation of line AB will be, which is parallel to CD, as opposite sides of parallelogram are parallel and equal,is equal to
2 x -3 y + k=0
Because when lines are parallel their slopes are equal.
→→Equation of line AD is , y= - 3 x.
Coordinates of point A can be calculated by
→2 x -3 × (-3 x ) +k=0
→2 x +9 x +k=0
![\rightarrow x=\frac{-k}{11}\\\\\rightarrow y=\frac{3k}{11}](https://tex.z-dn.net/?f=%5Crightarrow%20x%3D%5Cfrac%7B-k%7D%7B11%7D%5C%5C%5C%5C%5Crightarrow%20y%3D%5Cfrac%7B3k%7D%7B11%7D)
→→→Similarly, Coordinate of point D can be calculated by solving these two lines:
y = -3 x + 11
2 x -3 y + k=0
→2 x -3 × (-3 x +11) +k=0
→2 x +9 x -33 +k=0
→11 x =33 -k
![y=-3 \times \frac{33-k}{11}+11\\\\y=\frac{-99+3 k+121}{11}\\\\y=\frac{22+3 k}{11}](https://tex.z-dn.net/?f=y%3D-3%20%5Ctimes%20%5Cfrac%7B33-k%7D%7B11%7D%2B11%5C%5C%5C%5Cy%3D%5Cfrac%7B-99%2B3%20k%2B121%7D%7B11%7D%5C%5C%5C%5Cy%3D%5Cfrac%7B22%2B3%20k%7D%7B11%7D)
→→Coordinates of A is ![(\frac{-k}{11},\frac{3k}{11})](https://tex.z-dn.net/?f=%28%5Cfrac%7B-k%7D%7B11%7D%2C%5Cfrac%7B3k%7D%7B11%7D%29)
Coordinates of Point D is
.
you, can get infinite number of ordered pairs, for different value of k.
For, k=0 ,
A= (0,0)
D=(3,2)