Sum

The points A (2, 0), B (9, 1), C (11, 6) and D (4, 4) are the vertices of a quadrilateral ABCD. Determine whether ABCD is a rhombus or not.

Advertisement Remove all ads

#### Solution

The given points are A (2, 0), B (9, 1), C (11, 6) and D (4, 4).

Let us find the length of all the sides of the quadrilateral ABCD.

\[AB = \sqrt{\left( 2 - 9 \right)^2 + \left( 0 - 1 \right)^2} = \sqrt{50} = 5\sqrt{2}\]

\[BC = \sqrt{\left( 11 - 9 \right)^2 + \left( 6 - 1 \right)^2} = \sqrt{29}\]

\[CD = \sqrt{\left( 4 - 11 \right)^2 + \left( 4 - 6 \right)^2} = \sqrt{49 + 4} = \sqrt{53}\]

\[AD = \sqrt{\left( 4 - 2 \right)^2 + \left( 4 - 0 \right)^2} = \sqrt{4 + 16} = 2\sqrt{5}\]

\[\because AB \neq BC \neq CD \neq AD\], quadrilateral ABCD is not a rhombus.

Concept: Brief Review of Cartesian System of Rectanglar Co-ordinates

Is there an error in this question or solution?

Advertisement Remove all ads

#### APPEARS IN

Advertisement Remove all ads

Advertisement Remove all ads