Figure ABCD is transformed to figure A prime B prime C prime D prime, as shown below:

1 answer:

6 0

The sequence of transformations used to obtain figure A'B'C'D'from figure ABCD are a reflection across the y-axis followed by a horizontal translation 1 unit right

<h3>Complete question</h3>

Figure ABCD is transformed to figure A'B'C'D', as shown below: (see attachment)

Which of the following sequences of transformations is used to obtain figure A'B'C'D'from figure ABCD?

<h3>How to determine the transformation?</h3>

From the attached figure, we can see that:

ABCD and A'B'C'D' are on either sides of the y-axis.

This means that ABCD is first reflected across the y-axis.

ABCD is 2 units from the y-axis, while A'B'C'D' is 1 unit from the y-axis.

This means that the reflection is followed by a translation.

The unit of translation is:

Unit = 2 - 1

Evaluate

Unit = 1

1 represents a right translation

Hence, the sequence of transformations are a reflection across the y-axis followed by a horizontal translation 1 unit right

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Find parametric equations for the line through the point (0, 3, 2) that is parallel to the plane x + y + z = 5 and perpendicular

kozerog [31]

Hello :
let A(0,3,2) and (Δ) this line , v vector parallel to (<span>Δ).
M</span>∈ (Δ) : vector (AM) = t v..... t ∈ R

1 ) (Δ)  parallel to the plane x + y + z = 5 : let : n an vector <span>perpendicular
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2)  (Δ) perpendicular to the line (Δ') : x = 1+t , y = 3 - t , z = 2t :
vector (u) ⊥ v .... vector(u) parallel to (Δ') and vector(u) = (1 , -1 ,1)
vector (u) ⊥ vector (AM) means :
(1)(x)+(-1)(y-3)+(2)(z -2) =0
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so the system :
x+y+z - 5=0 ...(1)
x - y+2z - 1 = 0 ...(2)
(1)+(2) : 2x+3z - 6 =0
x = 3 - (3/2)z
subsct in (1) : 3 - (3/2)z +y +z - 5 =0
y = 1/2z +2
let : z=t
an parametric equations for the line (Δ) is : x = 3 - (3/2)t
  y = (1/2)t +2
z=t

verifiy :
1) (Δ)  parallel to the plane x + y + z = 5 :
(-3/2 , 1/2 ,1) <span>perpendicular to (1,1,1)
</span>because : (1)(-3/2)+(1)(1/2)+(1)(1) = -1 +1 = 0
2) (Δ) perpendicular to the line (Δ') :
(-3/2 , 1/2 ,1) perpendicular to (1,-1,2)
because : (1)(-3/2)+(-1)(1/2)+(1)(2) = -2 +2 = 0
A(0, 3, 2)∈(Δ) :
0 = 3-(3/2)t
3 = (1/2)t+2
2 =t
same : t = 2

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