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show that the pointsA 2,-1 , B 3,4 , c-2,3 , d -3,-2 are the verticles of a rhombus

Question

Show that the points A(2,-1), B(3,4), C(-2,3), D(-3,-2) are the vertices of a rhombus.

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Solution

To prove that the given points form a rhombus, we need to show that all sides are equal and diagonals are perpendicular to each other.

The points given are A(2,-1), B(3,4), C(-2,3), and D(-3,-2).

Step 1: Calculate the lengths of all sides using the distance formula: √[(x₂ - x₁)² + (y₂ - y₁)²]

AB = √[(3-2)² + (4-(-1))²] = √[1 + 25] = √26 BC = √[(-2-3)² + (3-4)²] = √[25 + 1] = √26 CD = √[(-3-(-2))² + (-2-3)²] = √[1 + 25] = √26 DA = √[(2-(-3))² + (-1-(-2))²] = √[25 + 1] = √26

So, AB = BC = CD = DA, which means all sides are equal.

Step 2: Calculate the lengths of the diagonals using the distance formula:

AC = √[(2-(-2))² + (-1-3)²] = √[16 + 16] = √32 BD = √[(3-(-3))² + (4-(-2))²] = √[36 + 36] = √72

Step 3: Check if the diagonals are perpendicular. If the product of their slopes is -1, then they are perpendicular.

Slope of AC = (3-(-1))/(-2-2) = -1 Slope of BD = (-2-4)/(-3-3) = 1

Since the product of the slopes of AC and BD is -1, the diagonals are perpendicular.

Therefore, the given points form a rhombus.

This problem has been solved

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