We are given a rectangle ABCD
A(-2, 3)
B(4, 6)
We are asked to find the slopes of sides BC, CD, and DA.
Let me first draw a rectangle to better understand the problem
Recall that the slope is given by
[tex]m=\frac{y_2−y_1}{ x_2−x_1}[/tex][tex]\text{where}(x_1,y_1)=(-2,3)\text{and}(x_2,y_2)=(4,6)[/tex]So the slope of side AB is
[tex]m_{AB}=\frac{6-3}{4-(-2)}=\frac{3}{4+2}=\frac{3}{6}=\frac{1}{2}=0.5[/tex]The sides BC and DA are perpenducluar to the side AB.
So their slopes will be
[tex]m_{BC}=m_{DA}=\frac{1}{-m_{AB}}[/tex]Substituting the value of slope of AB
[tex]m_{BC}=m_{DA}=\frac{1}{-0.5}=-2[/tex]The side CD is parallel to the side AB.
Parallel sides have equal slopes so
[tex]m_{CD}=m_{AB}=\frac{1}{2}[/tex]Therefore, the slopes of the rectangle ABCD are
[tex]\begin{gathered} m_{AB}=m_{CD}=\frac{1}{2} \\ m_{BC}=m_{DA}=-2 \end{gathered}[/tex]
