Question

show that the points a(3,1),b(0,-2),c(1,1),d(4,4),are the vertices of a parallelogram abcd

Answer 1

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Answer 2

Proved below.

Step-by-step explanation:

Given:

Let a(3, 1), b(0, -2), C(1, 1), D(4, 4) are the given points.

We know that , mid point of line segment joining the points $(x_1, y_1)$ and $(x_2,y_2)$ is $(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2} )$.

Now, co-ordinates of mid point of ac = $(\frac{3+1}{2},\frac{1+1}{2})$

                                                             = $(\frac{4}{2},\frac{2}{2})$

                                                              = (2, 1)                [1]

Co-ordinates of mid point of bd = $(\frac{0+4}{2} ,\frac{-2+4}{2})$

                                                       = (2, 1)                       [2]

From equation 1 and 2, co-ordinates of mid point of ac  = co-ordinates of mid point of bd

⇒ diagonals of quadrilateral abcd bisect each other.

⇒ abcd is a parallelogram

Hence proved.