Question

ABCD is a parallelogram with vertices A(x1,y1) , B(x2,y2) , C(x3,y3) . find the coordinates of the fourth vertex D in terms of x1 , x 2 , x3 , y1 , y2 and y3 .

Answer 1

We know that in the parallelogram the diagonals bisect each other:

A = (x₁,y₁)

B = (x₂,y₂)

C = (x₃,y₃)

D = (x,y)

Here the midpoint of the line would be AC = (x₁+x₃)/2

And the midpoint of the line BD = x₁ + x/2

x₁ + x₃/₂ = x+x₂/2

x = x₁+x₃+x₂

y₁ +y₃ = y₂ + y/2

y = y₁ + y₃ + y₂

So the coordinates for D will be (x₁+x₃+x₂,y₁+y₃+y₂)

Answer 2

Answer:

D(a,b)=D(x1+x3-x2 , y1+y3-y2).

Step-by-step explanation:

First draw the parallelogram with AB//CD and AD//BC.

Draw the diagonals AC and BD to intersect at O.

We know that diagonals of a parallelogram bisect each other at the point of intersection.

Therefore O is the midpoint of AC and BD.

Now find the co-ordinates of the point O by applying section formula on the diagonals AC (ratio in which O divides AC and BD is 1:1)

The co-ordinates come out to be O[(x1+x3)/2],[(y1+y3)/2].

Let us assume that co-ordinates of D is D(a,b).

Apply section formula on line segment BD and again find the co-ordinates of point O.

(Ratio is 1:1).

The co-ordinates come out to be O[(x2+a)/2],[(y2+b)/2].

Since the co-ordinates are of the same point O, they are equal.

Therefore,

O[(x1+x3)/2],[(y1+y3)/2]=O[(x2+a)/2],[(y2+b)/2]

Equate x coordinate and y coordinate separately.

The answer that follows is

D(a,b)=D(x1+x3-x2 , y1+y3-y2).