Consider the points A(3,−3), B(−2, 1), C(6, 0), and D(1, 4). Point A is joined to point B to create segment AB and point C is joined to point D to create segment CD.Part Aa. What is the slope of AB?b. What is the slope of CD?Segments AB and CD are translated 2 units to the left to get segments A′B′ and C′D′.Part Ba. What is the slope of A′B′?b. What is the slope of C′D′?

Answer:Part A:a. The slope of AB is [tex]-\frac{4}{5}[/tex]b. The slope of CD is [tex]-\frac{4}{5}[/tex]Part B:a. The slope of A'B' is [tex]-\frac{4}{5}[/tex]b. The slope of C'D' is [tex]-\frac{4}{5}[/tex]Step-by-step explanation:- The rule of a slope of a line whose endpoints are [tex](x_{1},y_{1})[/tex]   and [tex](x_{2},y_{2})[/tex] is [tex]m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}[/tex]Part A:∵ Point A = (3 , -3) and point B = (-2 , 1)∵ Point C = (6 , 0) and point D = (1 , 4)a.∴ [tex]m_{AB}=\frac{1-(-3)}{(-2)-3}=\frac{1+3}{-2-3}=\frac{4}{-5}[/tex]∴ The slope of AB is [tex]-\frac{4}{5}[/tex]b.∴ [tex]m_{CD}=\frac{4-0}{1-6}=\frac{4}{-5}[/tex]∴ The slope of CD is [tex]-\frac{4}{5}[/tex]- The image of point (x , y) by translation to the left k units is (x - k , y) - Translation doesn't change the slope of a line, so the line and its   image have same slopePart B:∵ Segments AB and CD are translated 2 units to the left to get    segments A′B′ and C′D′∵ Point A' = (3 - 2 , -3) and point B' = (-2 - 2 , 1)∴ Point A' = (1 , -3) and point B' = (-4 , 1)∵ Point C' = (6 - 2 , 0) and point D' = (1 - 2 , 4)∴ Point C' = (4 , 0) and point D' = (-1 , 4)a.∴ [tex]m_{A'B'}=\frac{1-(-3)}{(-4)-1}=\frac{1+3}{-4-1}=\frac{4}{-5}[/tex]∴ The slope of A'B' is [tex]-\frac{4}{5}[/tex]b.∴ [tex]m_{C'D'}=\frac{4-0}{-1-4}=\frac{4}{-5}[/tex]∴ The slope of C'D' is [tex]-\frac{4}{5}[/tex]