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Question
The three vertices of a parallelogram are (3, 4) (3, 8) and (9, 8). Find the fourth vertex.
Solution
Let A(3,4), B(3,8) and C(9,8) be the three given vertex then the fourth vertex D(x,y)
Since ABCD is a parallelogram, the diagonals bisect each other.
Therefore the mid-point of diagonals of the parallelogram coincide.
Let p(x,y) be the mid-point of diagonals AC then,
`P(x,y) = ((3 + 9)/2, (4 + 8)/2)`
P(x,y) = (6,6)
Let Q(x,y) be the mid point of diagonal BD. then
`Q(x,y) = ((3 + x)/2, (8 + y)/2)`
Coordinates of mid-point AC = Coordinates of mid-point BD
P(x,y) = Q(x,y)
`=> (6,6) = ((3 + x)/2, (8 + y)/2)`
Now equating individual components
`=> 6 = (3 + x)/2` and `6 = (8 + 4)/2`
=> 3 + x = 12 and 8 + y = 12
=> x = 9 and y = 4
Hence, coordinates of fourth points are (9, 4)
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