Question

.Prove that ABCD is rhombus , where A( 3, 10 ), B( -2, 6 ) , C(3, 2 ) and D( 8, 6)

Answer 1

Answer:

Step-by-step explanation:

For any quadrilateral to be a rhombus, all sides should be equal and diagonals are unequal and perpendicular bisectors of each other.

By applying distance formula,

=> AB =  $\sqrt{41}$

=> BC = $\sqrt41}$

=> CD = $\sqrt{41}$

=> DA = $\sqrt{41}$

Now, we will check for the diagonals,

=> AC = 8

=> BD = 10

NOW, to check if the diagonals are perpendicular or not, we can find their slopes.

=> Slope of AC = (10 - 2) / (3 -3) = 8 / 0 = Parallel to Y-axis

=> Slope of BD = (6 -6)/(8 +2) = 0/10 = Parallel to X-axis

Thus, we can say that the diagonals are perpendicular and hence all the conditions for a rhombus are satisfied.

HENCE, PROVED.

PLEASE MARK IT AS THE BRAINLIEST...

Answer 2

Step-by-step explanation:

For any quadrilateral to be a rhombus, all sides should be equal and diagonals are unequal and perpendicular bisectors of each other.

By applying distance formula,

=> AB = \sqrt{41}

41

=> BC = $$\sqrt41}$$

=> CD = $$\sqrt{41}$$

=> DA = $$\sqrt{41}$$

Now, we will check for the diagonals,

=> AC = 8

=> BD = 10

NOW, to check if the diagonals are perpendicular or not, we can find their slopes.

=> Slope of AC = (10 - 2) / (3 -3) = 8 / 0 = Parallel to Y-axis

=> Slope of BD = (6 -6)/(8 +2) = 0/10 = Parallel to X-axis

Thus, we can say that the diagonals are perpendicular and hence all the conditions for a rhombus are satisfied.

HENCE, PROVED.