Question

Show that points A (1, 1), B (-1, 5), C (7, 9) and D (9, 5) are the vertices of a rectangle ABCD

Answer 1

Answer:

∵ AB = CD

Hence, A(1, 1), B(-1, 5), C(7, 9) & D(9, 5) are the vertices of a rectangle ABCD.

Step-by-step explanation:

Given that:-

A = (1, 1), B = (-1, 5), C = (7, 9) & D = (9, 5)

Show that:-

A, B, C and D are the vertices of a rectangle ABCD.

We know that-

In a rectangle ABCD, the opposite sides are equal.

So, AB = CD

Using formula:-

Distance between two points P(x, y) and Q(a, b) is given by-

$PQ = \sqrt{(x-a)^{2}+(y-b)^{2}  }$

So, distance between A and B is-

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

$AB = \sqrt{(2)^{2}+(-4)^{2}}$

$AB = \sqrt{4+16}$

$AB = \sqrt{20}$

$AB = 2\sqrt{5}$

Similarly, distance between C(7, 9) and D(9, 5) is-

$CD = \sqrt{(7-9)^{2}+(9-5)^{2}}$

$CD = \sqrt{(-2)^{2}+(4)^{2}}$

$CD = \sqrt{4+16}$

$CD = \sqrt{20}$

$CD = 2\sqrt{5}$

∵ AB = CD

Hence, A(1, 1), B(-1, 5), C(7, 9) & D(9, 5) are the vertices of a rectangle ABCD.