Answer: (-4,-5)
Step-by-step explanation:
Here ABCD is a parallelogram,
Where A≡(-2,-1), B≡(2,1), C≡(0,-3) and D≡(x,y)
By the property of parallelogram,
AB║CD, AD║BC, AB = CD and AD=BC
If AB ║ CD
⇒ Slope of AB = Slope of CD
⇒ [tex]\frac{1-(-1)}{2-(-2)} = \frac{-3-y}{0-x}[/tex]
⇒ [tex]\frac{1+1}{2+2)} = \frac{-3-y}{-x}[/tex]
⇒ [tex]\frac{2}{4} = \frac{3+y}{x}[/tex]
⇒ [tex]x=6+2y[/tex] ------ (1)
Now, AB = CD
[tex]\sqrt{(2-(-2))^2+(1-(-1))^2} = \sqrt{(0-x)^2+(-3-y)^2}[/tex]
[tex]\sqrt{4^2+2^2} = \sqrt{x^2+9+6y+y^2}[/tex]
[tex]\sqrt{16+4} = \sqrt{x^2+9+6y+y^2}[/tex]
[tex]\sqrt{20} = \sqrt{x^2+9+6y+y^2}[/tex]
[tex]20 = x^2+9+6y+y^2[/tex]
[tex]x^2+6y+y^2=11[/tex]
From equation (1)
[tex](6+2y)^2+6y+y^2=11[/tex]
[tex]36+4y^2+24y+6y+y^2=11[/tex]
[tex]5y^2+30y+25=0[/tex]
[tex]y^2+6y+5=0[/tex]
⇒ y = -1 or -5
Again by equation (1)
for y = -1, x = 4
For y = -5, x = -4
Thus the coordinate of D are (4,-1) or ( -4,-5)
But for D≡(4,-1), AD∦BC
While For D≡(-4,-5) AD ║ BC
Thus the coordinates of D are ( -4,-5)