Question

part a
identify the figure with the vertices k(-1,1), l(3,4), m(6,0), and n(2,-3).
select choice

part b
find the perimeter and area of the figure.
perimeter: units
area: units²

Explanation:

Step1: Calculate side - lengths using distance formula

The distance formula between two points $(x_1,y_1)$ and $(x_2,y_2)$ is $d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$.
For side $KL$ with $K(-1,1)$ and $L(3,4)$:
$d_{KL}=\sqrt{(3 + 1)^2+(4 - 1)^2}=\sqrt{16 + 9}=\sqrt{25}=5$.
For side $LM$ with $L(3,4)$ and $M(6,0)$:
$d_{LM}=\sqrt{(6 - 3)^2+(0 - 4)^2}=\sqrt{9+16}=\sqrt{25}=5$.
For side $MN$ with $M(6,0)$ and $N(2,-3)$:
$d_{MN}=\sqrt{(2 - 6)^2+(-3 - 0)^2}=\sqrt{16 + 9}=\sqrt{25}=5$.
For side $NK$ with $N(2,-3)$ and $K(-1,1)$:
$d_{NK}=\sqrt{(-1 - 2)^2+(1 + 3)^2}=\sqrt{9 + 16}=\sqrt{25}=5$.

Step2: Calculate slopes to identify the figure

The slope formula between two points $(x_1,y_1)$ and $(x_2,y_2)$ is $m=\frac{y_2 - y_1}{x_2 - x_1}$.
Slope of $KL$, $m_{KL}=\frac{4 - 1}{3+1}=\frac{3}{4}$.
Slope of $LM$, $m_{LM}=\frac{0 - 4}{6 - 3}=-\frac{4}{3}$.
Since $m_{KL}\times m_{LM}=\frac{3}{4}\times(-\frac{4}{3})=- 1$, $KL\perp LM$.
Since all side - lengths are equal ($d_{KL}=d_{LM}=d_{MN}=d_{NK} = 5$) and adjacent sides are perpendicular, the figure is a square.

Step3: Calculate the perimeter

The perimeter $P$ of a square with side - length $s$ is $P = 4s$. Here $s = 5$, so $P=4\times5 = 20$ units.

Step4: Calculate the area

The area $A$ of a square with side - length $s$ is $A=s^2$. Here $s = 5$, so $A = 25$ units$^2$.

Answer:

Part A: Square
Part B:
perimeter: 20 units
area: 25 units$^2$