Question

Three vertices of a parallelogram ABCD taken in order A(3,6), B(5,10) and C(3,2) find;

the coordinates of the fourth vertex D

length of diagonal BD

equation of side AB of parallelogram ABCD

Answer 1

solution:

We have  A(3,6);B(5,10);C(3,2) are the given 3 vertices of parallelogram ABCD.

Let D(a,b) be the coordinates of the fourth vertex,

Now, we know that diagonals of a parallelogram bisect each other,

So, coordinates of mid point of BD = coordinates of mid point of AC.

 (5+a/2 ,10+b/2) = (3+3/2, 6+2/2)   using mid point formula x = x1+x2/2 : y = y1 +y2/2

 (5+a/2 ,10+b/2) = (3,4)

 5+a/2= 3 and 10+b/2 = 4

 5+a = 6 and 10+b = 8

 A = 1 and b = -2

So, coordinates of vertex d are : D(1,-2)

We know that distance between points (x1,y1) and (x2,y2) is

Distance = √(x2-x1)² + (y2-y1)²

Now, BD = √(5-1)²+ (10+2)2²

  = √16+144

   = √160 =4√10 units

We know that equation of line segment joining (x1,y1) and (x2,y2) is  

y-y1/x-x1 = y2-y1/x2-x1

so, equation of line segment joining A(3,6) and B(5,10) is  

y-6/x-3 = 10-6/5-3

 Y-6/x-3 = 4/2

 Y-6/x-3 = 2

 Y-6 = 2x -6

 Y-2x = 0

 Y = 2x