Question

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

1. the co-ordinates of the fourth vertex D. 
2. length of diagonal BD. 
3. equation of side AB of the parallelogram ABCD.

Answer 1

i. Let (x, y) be the co-ordinates of D. 

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

∴ Mid-point of diagonal AC = Mid-point of diagonal BD

`=> ((3 + 3)/2, (6 + 2)/2) = ((5 + x)/2, (10 + y)/2)` 

`=> (3, 4) = ((5 + x)/2, (10 + y)/2)`

`=> (5 + x)/2 = 3`

`=>` 5 + x = 6

`=>` x = 1

And `(10 + y)/2 = 4`

`=>` 10 + y = 8

`=>` y = –2

∴ Co-ordinates of D are (1, –2).

ii. Length of diagonal BD = `sqrt((1 - 5)^2 + (-2 - 10)^2)`

= `sqrt((-4)^2 + (-12)^2)`

= `sqrt(16 + 144)` 

= `sqrt(160)`

= `4sqrt(10)` units

iii. Slope of side AB = m = `(10 - 6)/(5 - 3) = 4/2 = 2`

Thus, the equation of side AB is given by 

y – 6 = 2(x – 3)

i.e. y – 6 = 2x – 6

i.e. 2x – y = 0

i.e 2x = y