Respuesta :
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 slope
Part 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]
Answer: The slopes of AB, CD, A'B' and C'D' are equal and is equal to [tex]-\dfrac{4}{5}.[/tex]
Step-by-step explanation: We are given 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.
We are given to answer the following two parts :
Part A : To find the slope of AB and CD.
We know that the slope of a line containing the points (a, b) and (c, d) is given by
[tex]m=\dfrac{d-b}{c-a}.[/tex]
Therefore, the slopes of AB and CD are
[tex]\textup{slope of AB}=\dfrac{1-(-3)}{-2-3}=\dfrac{4}{-5}=-\dfrac{4}{5},\\\\\\\textup{slope of CD}=\dfrac{4-0}{1-6}=\dfrac{4}{-5}=-\dfrac{4}{5}.[/tex]
Part B: Segments AB and CD are translated 2 units to the left to get segments A′B′ and C′D′.
To find the slopes of A'B' and C'D'.
We know that, if a point (x, y) is translated 2 units to the left, then its co-ordinates becomes
(x, y) ⇒ (x-2, y).
So, the co-ordinates of A', B', C' and D' are
A(3, -3) ⇒ A'(3-2, -3) = (1, -3),
B(-2, 1) ⇒ B'(-2-2, 1) = (-4, 1),
C(6, 0) ⇒ C'(6-2, 0) = (4, 0),
D(1, 4) ⇒ D'(1-2, 4) = (-1, 4).
Therefore, the slopes of A'B' and C'D' are
[tex]\textup{slope of A'B'}=\dfrac{1-(-3)}{-4-1}=-\dfrac{4}{5},\\\\\\\textup{slope of C'D'}=\dfrac{4-0}{-1-4}=-\dfrac{4}{5}.[/tex]
Thus, the slopes of AB, CD, A'B' and C'D' are equal and is equal to [tex]-\dfrac{4}{5}.[/tex]
