**(i)** Let the points (- 1, – 2), (1, 0), ( – 1, 2), and ( – 3, 0) be representing the vertices A, B, C, and D of the given quadrilateral respectively.

Side length = AB = BC = CD = DA = 2\( \sqrt{2}\)

Diagonal Measure = AC = BD = 4

Therefore, the given points are the vertices of a square.

**(ii)** Let the points (- 3, 5), (3, 1), (0, 3), and ( – 1, – 4) be representing the vertices A, B, C, and D of the given quadrilateral respectively.

Its also seen that points A, B and C are collinear. So, the given points can only form 3 sides i.e, a triangle and not a quadrilateral which has 4 sides.

Therefore, the given points cannot form a general quadrilateral.

**(iii)** Let the points (4, 5), (7, 6), (4, 3), and (1, 2) be representing the vertices A, B, C, and D of the given quadrilateral respectively.

Opposite sides of this quadrilateral are of the same length. However, the diagonals are of different lengths. Therefore, the given points are the vertices of a parallelogram.

Answered by Pragya Singh | 1 year agoIn the determine whether the given quadratic equations have real roots and if so, And the roots 3x^{2} – 2x + 2 = 0

Find the point on x-axis which is equidistant from the points (-2, 5) and (2, -3).

Answer the following questions:-

**(i) **Show that the points A (5, 6), B (1, 5), C (2, 1) and D (6, 2) are the vertices of a square.

**(ii)** Prove that the points A (2, 3), B (-2, 2), C (-1, -2) and D (3, -1) are the vertices of a square ABCD.

**(iii)** Name the type of triangle PQR formed by the point \( P(\sqrt{2} , \sqrt{2}), Q(- \sqrt{2}, – \sqrt{2)} and\; R (-\sqrt{6} , \sqrt{6} )\).

Find a point on the x-axis which is equidistant from the points (7, 6) and (-3, 4).

Find the coordinates of the circumcentre of the triangle whose vertices are (3, 0), (-1, -6) and (4, -1). Also, find its circumradius.