Question

NO LINKS!! Please help with questions 3 and 4​

Answer 1

`\textsf{3.} \quad (x,y) \rightarrow (x-5,y+2)`

`\textsf{4.} \quad (x,y) \rightarrow (x+3,y+1)`

Explanation:

Question 3

Given vertices of ΔLMN:

• L = (1, -1)
• M = (4, 1)
• N = (5, -1)

Given vertices of ΔL'M'N':

• L' = (-4, 1)
• M' = (-1, 3)
• N' = (0, 1)

From inspection of the given diagram, we can see that the two triangles are congruent since their corresponding angles and corresponding side lengths are the same.

There is no apparent reflection or rotation, so the transformation of ΔLMN to ΔL'M'N' is by translation.

To find the mapping rule for the translation, choose one pair of corresponding vertices and determine the difference between the x and y values of the translated point and the original point:

`x_(M')-x_M =-1-4=-5 \implies \textsf{5 units left}`

`y_(M')-y_M =3-1=2 \implies \textsf{2 units up}`

Therefore, the mapping rule for the translation is:

`(x,y) \rightarrow (x-5,y+2)`

Question 4

Given vertices of ΔLMN:

• L = (-6, -4)
• M = (-5, -1)
• N = (-2 -4)

Given vertices of ΔL'M'N':

• L' = (-3, -3)
• M' = (-2, 0)
• N' = (1, -3)

From inspection of the given diagram, we can see that the two triangles are congruent since their corresponding angles and corresponding side lengths are the same.

There is no apparent reflection or rotation, so the transformation of ΔLMN to ΔL'M'N' is by translation.

To find the mapping rule for the translation, choose one pair of corresponding vertices and determine the difference between the x and y values of the translated point and the original point:

`x_(L')-x_L =-3-(-6)=3 \implies \textsf{3 units right}`

`y_(L')-y_L =-3-(-4)=1 \implies \textsf{1 units up}`

Therefore, the mapping rule for the translation is:

`(x,y) \rightarrow (x+3,y+1)`

Answer 2

Answers:

Problem 3) transformation is
`(\text{x},\text{y})\to(\text{x}-5,\text{y}+2)`

Problem 4) transformation is
`(\text{x},\text{y})\to(\text{x}+3,\text{y}+1)`

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Explanation for problem 3

preimage point L is at (1, -1)

image point L' is at (-4, 1)

When going from x = 1 to x = -4, we shift left 5 units. So we'll have x-5 as part of the transformation.

When going from y = -1 to y = 1, we'll shift 2 units up and involve y+2

Since we have x become x-5, and y become y+2, the transformation is
`(\text{x},\text{y})\to(\text{x}-5,\text{y}+2)\\\\`

Let's test the transformation on point L(1,-1)

`(\text{x},\text{y})\to(\text{x}-5,\text{y}+2)\\\\(1,-1)\to(1-5,-1+2)\\\\(1,-1)\to(-4,1)\\\\`

which helps confirm we have the correct transformation. I'll let you check the other points M and N.

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Explanation for problem 4

• L has an x coordinate of x = -6
• L' has an x coordinate of x = -3

When going from L to L', we need to shift 3 units to the right. Each x becomes x+3

• L has a y coordinate of y = -4
• L' has a y coordinate of y = -3

This tells us to shift up 1 unit. The y coordinate becomes y+1

So we can say

`\text{x} \to \text{x}+3\\\\\text{y} \to \text{y}+1`

which condenses to the notation

`(\text{x},\text{y})\to(\text{x}+3,\text{y}+1)\\\\`

Let's test this transformation on point M(-5,-1) and we should get to M'(-2,0)

`(\text{x},\text{y})\to(\text{x}+3,\text{y}+1)\\\\(-5,-1)\to(-5+3,-1+1)\\\\(-5,-1)\to(-2,0)\\\\`

This confirms the translation rule works for point M.

I'll let you check the other points L and N.