Question

Figure LMNO is located at L (1, 1), M (2, 3), N (4, 3), and O (5, 1). Use coordinate geometry to best classify quadrilateral LMNO. Kite Parallelogram Square Trapezoid

Answer 1

Answer:

Quadrilateral LMNO is a trapezium.

Step-by-step explanation:

Given there are four coordinates:

L (1, 1), M (2, 3), N (4, 3), and O (5, 1)

Please refer to attached image.

To determine the type of quadrilateral, we need to see the sides and their slopes.

1. Parallel sides have equal slopes. (Helpful in Square, parallelogram and trapezium)

2. Multiplication of slopes of two perpendicular lines is equal to -1. (helpful in square in which adjacent sides are perpendicular to each other)

Now, let us find distance of each side using distance formula:

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

$LM = \sqrt{(2-1)^2+(3-1)^2}\\LM =\sqrt{1+4}\\LM =\sqrt{5}$

$MN = \sqrt{(4-2)^2+(3-3)^2}\\MN =\sqrt{4+0}\\MN =2$

$NO= \sqrt{(5-4)^2+(1-3)^2}\\NO=\sqrt{1+4}\\NO=\sqrt{5}$

$OL = \sqrt{(5-1)^2+(1-1)^2}\\OL =\sqrt{16+0}\\OL=4$

Only two sides are equal to each other, so it can not be square or parallelogram.

Now, let us have a look at the slopes of line:

Slope is given as:

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

$m_{LM} = \dfrac{3-1}{2-1} = \dfrac{2}{1} = 2$

$m_{MN} = \dfrac{3-3}{4-2} = \dfrac{0}{2} = 0$

$m_{NO} = \dfrac{1-3}{5-4} = \dfrac{-2}{1} = -2$

$m_{OL} = \dfrac{1-1}{5-1} = \dfrac{0}{4} = 0$

Slope of two lines is equal i.e. 2 lines are parallel.

Hence, the given quadrilateral is a trapezium.

Answer 2

Answer:

The answer is  " Trapezoid"

Step-by-step explanation:

I got it right on my test