Question

The quadrilateral ABCD on a coordinate plane has the following characteristics.

Answer 1

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

  $\frac{y-5}{x-2}=\frac{3-5}{-1-2}\\\\-3 \times (y-5)=-2\times (x-2)\\\\-3 y +15=-2 x +4\\\\2 x -3 y +11=0$

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

$\rightarrow x=\frac{-k}{11}\\\\\rightarrow y=\frac{3k}{11}$

→→→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

$x=\frac{33-k}{11}$

$y=-3 \times \frac{33-k}{11}+11\\\\y=\frac{-99+3 k+121}{11}\\\\y=\frac{22+3 k}{11}$

→→Coordinates of A is $(\frac{-k}{11},\frac{3k}{11})$

Coordinates of Point D is $(\frac{33-k}{11},\frac{22+3k}{11})$.

you, can get infinite number of ordered pairs, for different value of k.

For, k=0 ,

A= (0,0)

D=(3,2)