Question

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 A
a. 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 B
a. What is the slope of A′B′?
b. What is the slope of C′D′?

Answer 1

Answer:

Part A:

a. The slope of AB is $-\frac{4}{5}$

b. The slope of CD is $-\frac{4}{5}$

Part B:

a. The slope of A'B' is $-\frac{4}{5}$

b. The slope of C'D' is $-\frac{4}{5}$

Step-by-step explanation:

- The rule of a slope of a line whose endpoints are $(x_{1},y_{1})$

  and $(x_{2},y_{2})$ is $m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

Part A:

∵ Point A = (3 , -3) and point B = (-2 , 1)

∵ Point C = (6 , 0) and point D = (1 , 4)

a.

∴ $m_{AB}=\frac{1-(-3)}{(-2)-3}=\frac{1+3}{-2-3}=\frac{4}{-5}$

∴ The slope of AB is $-\frac{4}{5}$

b.

∴ $m_{CD}=\frac{4-0}{1-6}=\frac{4}{-5}$

∴ The slope of CD is $-\frac{4}{5}$

- 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.

∴ $m_{A'B'}=\frac{1-(-3)}{(-4)-1}=\frac{1+3}{-4-1}=\frac{4}{-5}$

∴ The slope of A'B' is $-\frac{4}{5}$

b.

∴ $m_{C'D'}=\frac{4-0}{-1-4}=\frac{4}{-5}$

∴ The slope of C'D' is $-\frac{4}{5}$

Answer 2

Answer:  The slopes of AB, CD, A'B' and C'D' are equal and is equal to $-\dfrac{4}{5}.$

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

$m=\dfrac{d-b}{c-a}.$

Therefore, the slopes of AB and CD are

$\textup{slope of AB}=\dfrac{1-(-3)}{-2-3}=\dfrac{4}{-5}=-\dfrac{4}{5},\\\\\\\textup{slope of CD}=\dfrac{4-0}{1-6}=\dfrac{4}{-5}=-\dfrac{4}{5}.$

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

$\textup{slope of A'B'}=\dfrac{1-(-3)}{-4-1}=-\dfrac{4}{5},\\\\\\\textup{slope of C'D'}=\dfrac{4-0}{-1-4}=-\dfrac{4}{5}.$

Thus, the slopes of AB, CD, A'B' and C'D' are equal and is equal to $-\dfrac{4}{5}.$