Let the given point be A (-2, -1) , B (4, 0) , C ( 3, 3) and D ( -3, 2)

So now, The slope of AB = \( \dfrac{(0+1)}{(4+2)}\)

= \(\dfrac{1}{6}\)

The slope of CD

= \( \dfrac{(3-2)}{(3+3)}\) = \( \dfrac{1}{6}\)

Hence, slope of AB = Slope of CD

AB ∥ CD

Now,

The slope of BC = \( \dfrac{(3-0)}{(3-4)}\)

= \( \dfrac{3}{-1}\) = -3

The slope of AD =\( \dfrac{(2+1)}{(-3+2)}\)

= \( \dfrac{3}{-1}\) = -3

Hence, slope of BC = Slope of AD

BC ∥ AD

Thus the pair of opposite sides are quadrilateral are parallel, so we can say that ABCD is a parallelogram.

Hence the given vertices, A (-2, -1), B (4, 0), C(3, 3) and D(-3, 2) are vertices of a parallelogram

Answered by Pragya Singh | 1 year agoFind the angle between the line joining the points (2, 0), (0, 3) and the line x + y = 1.

Prove that the points (2, -1), (0, 2), (2, 3) and (4, 0) are the coordinates of the vertices of a parallelogram and find the angle between its diagonals.

Find the acute angle between the lines 2x – y + 3 = 0 and x + y + 2 = 0.

Find the angles between pairs of straight lines 3x – y + 5 = 0 and x – 3y + 1 = 0

Find the angles between pairs of straight lines 3x + y + 12 = 0 and x + 2y – 1 = 0