Answer
[tex]\begin{equation*} Slope\text{ of AD }=\frac{2k}{2j-b} \end{equation*}[/tex][tex]Slope\text{ of BC}=\frac{2k}{2j-b}[/tex]
AD || BC
Slope of AB = 0
Slope of DC = 0
AB || DC
Step-by-step explanation
The slope, m, of the line that passes through the points (x₁, y₁) and (x₂, y₂) is calculated as follows:
[tex]m=\frac{y_2-y_1}{x_2-x_1}[/tex]
Line AD passes through A(0, 0) and D(2j-b, 2k), then its slope is:
[tex]\begin{gathered} Slope\text{ of AD }=\frac{2k-0}{2j-b-0} \\ Slope\text{ of AD }=\frac{2k}{2j-b} \end{gathered}[/tex]
Line BC passes through B(b, 0) and C(2j, 2k), then its slope is:
[tex]\begin{gathered} Slope\text{ of BC }=\frac{2k-0}{2j-b} \\ Slope\text{ of BC}=\frac{2k}{2j-b} \end{gathered}[/tex]
Given that sides AD and BC have the same slope, then they are parallel, that is,
[tex]\bar{AD}||\bar{BC}[/tex]
Line AB passes through A(0, 0) and B(b, 0), then its slope is:
[tex]\begin{gathered} slope\text{ of AB }=\frac{0-0}{b-0} \\ slope\text{ of AB }=\frac{0}{b} \\ slope\text{ of AB}=0 \end{gathered}[/tex]
Line DC passes through D(2j-b, 2k) and C(2j, 2k), then its slope is:
[tex]\begin{gathered} slope\text{ of DC }=\frac{2k-2k}{2j-(2j-b)} \\ slope\text{ of DC }=\frac{0}{b} \\ slope\text{ of DC }=0 \end{gathered}[/tex]
Given that sides AB and DC have the same slope, then they are parallel, that is,
[tex]\bar{AB}||\bar{DC}[/tex]
