Question

Classifying parallelograms in the coordinate planePart B:Slope of KLSlope of side adjacent to KLPart c:From parts (a) and (b), what can we conclude about parallelogramJKLM? Check all that apply.O JKL M is a rectangle.O JKL M is a rhombus.O JKL M is a square.O JKL M is none of these

Answer 1

Given

J(2, -7) ; K(-6, -2) ; L(-1, 6) ; M(7, 1 )

Find

Length of KL and length of side adjacent to KL

Explanation

by distance formula we find the length of sides.

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

so ,

$\begin{gathered} JK=\sqrt{(-6-2)^2+(-2+7)^2}=\sqrt{64+25}=\sqrt{89} \\ KL=\sqrt{(-1+6)^2+(6+2)^2}=\sqrt{25+64}=\sqrt{89} \\ LM=\sqrt{(7+1)^2+(1-6)^2}=\sqrt{25+64}=\sqrt{89} \\ MJ=\sqrt{(7-2)^2+(1+7)^2}=\sqrt{25+64}=\sqrt{89} \\ JL=\sqrt{(-1-2)^2+(6+7)^2}=\sqrt{9+169}=\sqrt{178} \\ KM=\sqrt{(7+6)^2+(1+2)^2}=\sqrt{169+9}=\sqrt{178} \end{gathered}$

here all sides are equal and diagonal are equal , so it is a square .

a) Length of KL and length of side adjacent to KL is

$\sqrt{89}$

slope of KL is given by

$\begin{gathered} \frac{y_2-y_1}{x_2-x_1} \\ \\ \frac{6+2}{-1+6} \\ \frac{8}{5} \end{gathered}$

slope of side adjacent to KL is given by

$\begin{gathered} \frac{1-6}{7+1} \\ -\frac{5}{8} \end{gathered}$

Final Answer

a) Length of KL and side adjacent to KL is

$\sqrt{89}$

b) Slope of KL = 8/5

slope of side adjacent to KL is -5/8

c) it is a sqaure.