Question

PLEASE HELP!!!!!

Answer 1

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

⇒ $\frac{1-(-1)}{2-(-2)} = \frac{-3-y}{0-x}$

⇒ $\frac{1+1}{2+2)} = \frac{-3-y}{-x}$

⇒ $\frac{2}{4} = \frac{3+y}{x}$

⇒ $x=6+2y$ ------ (1)

Now, AB = CD

$\sqrt{(2-(-2))^2+(1-(-1))^2} = \sqrt{(0-x)^2+(-3-y)^2}$

$\sqrt{4^2+2^2} = \sqrt{x^2+9+6y+y^2}$

$\sqrt{16+4} = \sqrt{x^2+9+6y+y^2}$

$\sqrt{20} = \sqrt{x^2+9+6y+y^2}$

$20 = x^2+9+6y+y^2$

$x^2+6y+y^2=11$

From equation (1)

$(6+2y)^2+6y+y^2=11$

$36+4y^2+24y+6y+y^2=11$

$5y^2+30y+25=0$

$y^2+6y+5=0$

⇒ 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)