Question

ertices of a parallelogram ABCD taken in order are A(3, 6), B(5, 10) and C(3
(1) the co-ordinates of the fourth vertex D.
2. length of diagonal BD.
3. equation of side AB of the parallelogram ABCD.
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Answer 1

Answer:

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

Hope it helps you .....