P id the mid-point of side AB,

Coordinate of P = \( \dfrac{ (-1 – 1)}{2}= \dfrac{ (-1 + 4)}{2}= -1, \dfrac{3}{2}\)

Similarly, Q, R and S are (As Q is mid-point of BC, R is midpoint of CD and S is midpoint of AD)

Coordinate of Q = (2, 4)

Coordinate of R = (5,\( \dfrac{3}{2}\))

Coordinate of S = (2, -1)

Now,

Length of PQ = \( \sqrt{(-1 – 2)}^2 + ( \dfrac{3}{2} – 4)^2 = \dfrac{\sqrt{61}}{4}= \dfrac{\sqrt{61}}{2}\)

Length of SP = \( \sqrt{(2+1)}^2 + (-1- \dfrac{3}{2})^2 = \dfrac{\sqrt{61}}{4}= \dfrac{\sqrt{61}}{2}\)

Length of QR = \( \sqrt{(2-5)}^2 + (4- \dfrac{3}{2} )^2 = \dfrac{\sqrt{61}}{4}= \dfrac{\sqrt{61}}{2}\)

Length of RS = \( \sqrt{(5– 2)}^2 + ( \dfrac{3}{2} +1)^2 = \dfrac{\sqrt{61}}{4}= \dfrac{\sqrt{61}}{2}\)

Length of PR (diagonal) = \( \sqrt{(-1 – 5)}^2 + ( \dfrac{3}{2} – \dfrac{3}{2} )^2 =6\)

Length of QS (diagonal) = \( \sqrt{(2– 2)}^2 + (4+1 )^2 =5\)

The above values show that, PQ = SP = QR = RS =\( \dfrac{\sqrt{61}}{2}\), i.e. all sides are equal.

But PR ≠ QS i.e. diagonals are not of equal measure.

Hence, the given figure is a rhombus.

Answered by Abhisek | 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.