PLEASE HELP!!!!!
If you are given the 3 vertices of a parallelogram below, find the exact coordinates of the 4th vertex. Please find the coordinates for point D:

A (-2, -1)

B (2, 1)

C (0, -3)

D (x, y)

Please explain all the steps and show all work!

Respuesta :

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)

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