Respuesta :
For two lines to be parallel,
[tex]\begin{gathered} m_1=m_2 \\ m_1=Gradient\text{ of line AB} \\ m_2=\text{Gradient of line CD} \end{gathered}[/tex]For two lines to be perpendicular,
[tex]m_1\times m_2=-1[/tex]The formula for the gradient of a line passing through points (x1,y1) and (x2,y2) is given as
[tex]m=\frac{y_2-y_1}{x_2-x_1}[/tex]For line AB, the point given is
[tex]\begin{gathered} (-3,8)\text{ and (3,2)} \\ x_1=-3,y_1=8,x_2=3,y_2=2 \end{gathered}[/tex]By substituting in the formula, we will have
[tex]\begin{gathered} m=\frac{2-8}{3-(-3)} \\ m=-\frac{6}{3+3} \\ m_1=-\frac{6}{6} \\ m_1=-1 \end{gathered}[/tex]For line CD, the point given is
[tex]\begin{gathered} (7,1)\text{ and (5,-1)} \\ x_1=7,y_1=1,x_2=5,y_2=-1 \end{gathered}[/tex]By substituting the values in the formula, we will have
[tex]\begin{gathered} m_2=\frac{-1-1}{5-7} \\ m_2=-\frac{2}{-2} \\ m_2=1 \end{gathered}[/tex][tex]\begin{gathered} m_1\ne m_2 \\ \text{therefore the two lines are not parallel} \end{gathered}[/tex][tex]\begin{gathered} m_1\times m_2 \\ =-1\times1 \\ =-1 \end{gathered}[/tex]Therefore,
[tex]\begin{gathered} \sin ce \\ m_1\times m_2=-1 \\ \text{Then the two lines are perpendicular} \end{gathered}[/tex]Hence,
Line AB and line CD are perpendicular
