Question

Which statement is true about quadrilateral ABCD with vertices A(2, 8), B(3, 11), C(4, 8), and D(3, 5)?

Answer 1

Answer:

The quadrilateral is a rhombus

Step-by-step explanation:

Given

$A = (2, 8)$

$B = (3, 11)$

$C = (4, 8)$

$D=(3, 5)$

Required

The true statement

Calculate slope (m) using

$m = \frac{y_2 - y_1}{x_2 - x_1}$

Calculate distance using:

$d= \sqrt{(x_2 - x_1)^2 + (y_2 -y_1)^2}$

Calculate slope and distance AB

$m_{AB} = \frac{11 - 8}{3 - 2}$

$m_{AB} = \frac{3}{1}$

$m_{AB} = 3$ -- slope

$d_{AB}= \sqrt{(3 - 2)^2 + (11 -8)^2}$

$d_{AB}= \sqrt{10}$ -- distance

Calculate slope and distance BC

$m_{BC} = \frac{8 - 11}{4 - 3}$

$m_{BC} = \frac{- 3}{1}$

$m_{BC} = -3$ -- slope

$d_{BC} = \sqrt{(4-3)^2+(8-11)^2$

$d_{BC} = \sqrt{10}$ --- distance

Calculate slope CD

$m_{CD} = \frac{5 - 8}{3 - 4}$

$m_{CD} = \frac{- 3}{- 1}$

$m_{CD} = 3$ -- slope

$d_{CD} = \sqrt{(3-4)^2+(5-8)^2}$

$d_{CD} = \sqrt{10}$ -- distance

Calculate slope DA

$m_{DA} = \frac{8 - 5}{2 - 3}$

$m_{DA} = \frac{3}{- 1}$

$m_{DA} = -3$ -- slope

$d_{DA} = \sqrt{(2-3)^2 + (8-5)^2}$

$d_{DA} = \sqrt{10}$

From the computations above, we can see that all 4 sides are equal, i.e. $\sqrt{10}$

And the slope of adjacent sides are negative reciprocal, i.e.

$m_{AB} = 3$  and $m_{CD} = -3$

$m_{CD} = 3$ and $m_{DA} = -3$

The quadrilateral is a rhombus