For 11 and 12, find... a.) The slope of the line parallel to the line that passes through the given points, and b.) The slope of the line perpendicular to the line that passes through the given points. 11. ) (0, 3) and (8, 7) 12.) (5, -1) and (3, 1)

(11)
we are given two points
(x1,y1)=(0,3)
(x2,y2)=(8,7)
so, we get
[tex]x_1=0,y_1=3[/tex]
[tex]x_2=8,y_2=7[/tex]
(a)
we know that slope of parallel lines are always equal
so, we can use formula
[tex]m=\frac{y_2-y_1}{x_2-x_1}[/tex]
now, we can plug values
[tex]m=\frac{7-3}{8-0}[/tex]
[tex]m=\frac{4}{8}[/tex]
[tex]m=\frac{1}{2}[/tex]............Answer
(b)
we know that slope of perpendicular line is -1/m
so,
[tex]slope=\frac{-1}{m}[/tex]
now, we can plug value
[tex]slope=\frac{-1}{\frac{1}{2}}[/tex]
[tex]slope=-2[/tex]...............Answer
(12)
we are given two points
(x1,y1)=(5,-1)
(x2,y2)=(3,1)
so, we get
[tex]x_1=5,y_1=-1[/tex]
[tex]x_2=3,y_2=1[/tex]
(a)
we know that slope of parallel lines are always equal
so, we can use formula
[tex]m=\frac{y_2-y_1}{x_2-x_1}[/tex]
now, we can plug values
[tex]m=\frac{1+1}{3-5}[/tex]
[tex]m=\frac{2}{-2}[/tex]
[tex]m=-1[/tex]............Answer
(b)
we know that slope of perpendicular line is -1/m
so,
[tex]slope=\frac{-1}{m}[/tex]
now, we can plug value
[tex]slope=\frac{-1}{-1}[/tex]
[tex]slope=1[/tex]...............Answer