Question

what type of quadrilateral has the vertices a(3, 6), b(3, 3), c(6, 3), and d(6, 6)?

a. square

b. non - square parallelogram

c. non - square rhombus

d. non - special parallelogram

Explanation:

Step1: Calculate side lengths

Use the distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.

• For $AB$: $A(3,6)$, $B(3,3)$

$d_{AB} = \sqrt{(3 - 3)^2 + (3 - 6)^2} = \sqrt{0 + 9} = 3$

• For $BC$: $B(3,3)$, $C(6,3)$

$d_{BC} = \sqrt{(6 - 3)^2 + (3 - 3)^2} = \sqrt{9 + 0} = 3$

• For $CD$: $C(6,3)$, $D(6,6)$

$d_{CD} = \sqrt{(6 - 6)^2 + (6 - 3)^2} = \sqrt{0 + 9} = 3$

• For $DA$: $D(6,6)$, $A(3,6)$

$d_{DA} = \sqrt{(3 - 6)^2 + (6 - 6)^2} = \sqrt{9 + 0} = 3$

Step2: Check angles (slopes for perpendicularity)

Slope formula: $m = \frac{y_2 - y_1}{x_2 - x_1}$.

• Slope of $AB$: $m_{AB} = \frac{3 - 6}{3 - 3}$ (undefined, vertical line)
• Slope of $BC$: $m_{BC} = \frac{3 - 3}{6 - 3} = 0$ (horizontal line)

Vertical and horizontal lines are perpendicular (angle = $90^\circ$).
Similarly, $BC \perp CD$ (horizontal $\perp$ vertical), $CD \perp DA$ (vertical $\perp$ horizontal), $DA \perp AB$ (horizontal $\perp$ vertical).

All sides are equal ($3$ units) and all angles are right angles. Thus, it is a square.

Answer:

A. square