Question

question 4 of 5 select the correct response from each dropdown menu. quadrilateral jklm has vertices of j(0,0), k(0,a), l(4a,4a), m(a,0). what are the steps to complete the proof to show jklm is a kite? quadrilateral jklm has vertices of j(0,0), k(0,a), l(4a,4a), m(a,0) is given. using the distance formula, the length of kj and mj units, and the length of kl and ml is units. the slope , and the slope of jl is . since adj equal, then the quadrilateral jklm is a kite.

Explanation:

Step1: Recall 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 $\overline{KJ}$ with $K(0,a)$ and $J(0,0)$:
$d_{KJ}=\sqrt{(0 - 0)^2+(a - 0)^2}=a$.
For $\overline{MJ}$ with $M(a,0)$ and $J(0,0)$:
$d_{MJ}=\sqrt{(a - 0)^2+(0 - 0)^2}=a$.

Step2: Calculate length of $\overline{KL}$ and $\overline{ML}$

For $\overline{KL}$ with $K(0,a)$ and $L(4a,4a)$:
$d_{KL}=\sqrt{(4a - 0)^2+(4a - a)^2}=\sqrt{(4a)^2+(3a)^2}=\sqrt{16a^{2}+9a^{2}}=\sqrt{25a^{2}} = 5a$.
For $\overline{ML}$ with $M(a,0)$ and $L(4a,4a)$:
$d_{ML}=\sqrt{(4a - a)^2+(4a - 0)^2}=\sqrt{(3a)^2+(4a)^2}=\sqrt{9a^{2}+16a^{2}}=\sqrt{25a^{2}} = 5a$.

Step3: Recall slope - formula

The slope formula between two points $(x_1,y_1)$ and $(x_2,y_2)$ is $m=\frac{y_2 - y_1}{x_2 - x_1}$.
The slope of $\overline{JL}$ with $J(0,0)$ and $L(4a,4a)$ is $m_{JL}=\frac{4a - 0}{4a - 0}=1$.

Answer:

The length of $\overline{KJ}$ and $\overline{MJ}$ is $a$ units, and the length of $\overline{KL}$ and $\overline{ML}$ is $5a$ units. The slope of $\overline{JL}$ is $1$.