Question

Show that points A(-4, -7), B(-1, 2), C(8,5) and D(5, -4) are vertices of a
rhombus ABCD.​

Answer 1

Answer:

Step-by-step explanation:

Given,

A(-4 , -7)

B(-1 , 2)

C(8 , 5)

D(5 , -4)

To Show :-

These are the vertices of Rhombus.

Solution :-

To show the above co-ordinates are the vertices of Rhombus we need to prove that length of all sides are equal.

that means we need to prove that :-

AB = BC = CD = DA

Distance formula :-

$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

for AB :-

$x_2 = - 1 , y_2 = 2\\x_1 = - 4 , y_1 = - 7$

$d = \sqrt{(-1+4)^2 + (2 + 7)^2}$

$= \sqrt{(3)^2+(9)^2}$

$= \sqrt{9 + 81}$

$= \sqrt{90}$

Since, distance b/w AB = $\sf\sqrt{90}$

for BC :-

$x_2 = 8,y_2 = 5\\x_1 = -1,y_1 = 2$

$d = \sqrt{(8 + 1)^2 + (5 - 2)^2}$

$= \sqrt{(9)^2 + (3)^2}$

$= \sqrt{81 + 9}$

$= \sqrt{90}$

Since, distance b/w BC = $\sf\sqrt{90}$

for CD :-

$x_2 = 5,y_2 = -4\\x_1 = 8 , y_1 = 5$

$d = \sqrt{(5-8)^2+(-4-5)^2}$

$= \sqrt{(-3)^2 + (-9)^2}$

$= \sqrt{9 + 81}$

$= \sqrt{90}$

since, distance b/w CD =  $\sf\sqrt{90}$

for DA :-

$x_2 = -4,y_2 = -7\\x_1 = 5,y_1 = -4$

$d = \sqrt{(-4-5)^2 + (-7 + 4)^2}$

$= \sqrt{(-9)^2 + (-3)^2}$

$= \sqrt{81 + 9}$

$= \sqrt{90}$

since, distance b/w DA =  $\sf\sqrt{90}$

Since , we have proved that :-

AB = BC = CD = DA

Hence these are the vertices of Rhombus.

Answer 2

Answer:

I hope its helpful for u thx for your points