Question

The coordinates of quadrilateral ABCD are A(-1,-5), B(8,2), C(11,13), and D(2,6). Using coordinate geometry, prove that quadrilateral ABCD is a rhombus

Answer 1

Answer:

The quadrilateral ABCD is a rhombus.

Step-by-step explanation:

A rhombus is a quadrilateral having equal sides.

If ABCD is a rhombus then,

AB = BC = CD =  DA

It is provided that the coordinates of the rhombus ABCD are:

A = (-1, -5)

B = (8, 2)

C = (11, 13)

D = (2, 6)

Use the distance formula to compute the lengths of AB, BC, CD and DA.

The distance formula is:

$d=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$

Compute the length of AB:

$AB=\sqrt{(8-(-1))^{2}+(2-(-5))^{2}}=\sqrt{130}=11.4$

Compute the length of BC:

$BC=\sqrt{(11-8)^{2}+(13-2)^{2}}=\sqrt{130}=11.4$

Compute the length of CD:

$CD=\sqrt{(2-11)^{2}+(6-13)^{2}}=\sqrt{130}=11.4$

Compute the length of DA:

$DA=\sqrt{(2-(-1))^{2}+(6-(-5))^{2}}=\sqrt{130}=11.4$

Thus, the lengths AB, BC, CD and DA are equal, i.e. all sides are of length 11.4.

Hence proved that the quadrilateral ABCD is a rhombus.