Quadrilateral ABCD has vertices at A0, 6 , B4, 1 , C4, 0 and D8, 7 . Prove that: (a) ABCD is a rhombus using the distance formula (b) The diagonals of ABCD are perpendicular

Question

Quadrilateral ABCD has vertices at A0, 6 , B4, 1 , C4, 0 and D8, 7 . Prove that: (a) ABCD is a rhombus using the distance formula (b) The diagonals of ABCD are perpendicular

Answer 1

Jasmina,

The question is unclear but I read the 4 points as: A (0, 6), B (4, 1), C (-4, 0), D (-8, 7).

Those 4 points, ABCD do form a convex quadrilateral.

But the 2 pairs of opposite sides are then not parallel, so it's not a rhombus!

Here's how to tackle. it.

(a) Find the slope of AB, using (y_B - y_A)/(x_B - x_A). Do the same for CD. The 2 slopes should be equal, meaning AB || CD.

Do the same for AD and BC. They should also have the same slopes , so AD || BC.

Therefore, ABCD is a parallelogram.

To be a rhombus (a squashed over square), the 4 sides must also be equal in length.

Use the distance formula L = √((x₁² - x₂²) + (y₁² - y₂²)) to calculate the lengths of the above pairs of sides. They should all be equal => ABCD is a rhombus.

(b) To show that the diagonals, AC and BD, are perpendicular, calculate their slopes. m_AC and m_BD.

If m_AC x m_BD = -1, then AC is ⊥ BD

Calculate m_AC and m_BD by (y_C - y_A) / (x_C - x_A) and (y_B - y_A / (x_B - x_D)

Check that m_AC x m_BD = -1 which proves that AC ⊥ BD, as required.