Answer:
The vertices of the triangle after the translation are (-1 , -1) , (-4 , -5) , (-8 , 0)
The image of Δ NPQ after the translation is Δ KLM
Explanation:
*Lets revise translation
- If point (x , y) translate to the right h units
∴ Its image is (x + h , y)
- If point (x , y) translate to left h units
∴ Its image is (x - h , y)
- If point (x , y) translate up k units
∴ Its image is (x , y + k)
- If point (x , y) translate down k units
∴ Its image is (x , y - k)
* Now lets solve the problem
- From the graph the vertices of the triangle are
(0 , 2) , (-3 , -2) , (-7 , 3)
- The triangle will translate by the rule (x , y) ⇒ (x - 1 , y - 3)
∴ The triangle translate 1 unit to the left and 3 units down
- We will subtract each x-coordinate by 1 and each y-coordinate by 3
∴ The image of point (0 , 2) is (0 -1 , 2 - 3) = (-1 , -1)
∴ The image of point (-3 , -2) is (-3 - 1 , -2 - 3) = (-4 , -5)
∴ The image of point (-7 , 3) is (-7 - 1 , 3 - 3) = (-8 , 0)
* The vertices of the triangle after the translation are
(-1 , -1) , (-4 , -5) , (-8 , 0)
- From the graph the vertices of the triangle NPQ are
N (-7 , -6) , P (-4 , -3) , Q (-4 , -6)
- The triangle will translate by the rule (x , y) ⇒ (x + 8 , y + 1)
∴ The triangle translate 8 units to the right and 1 unit up
- We will add each x-coordinate by 8 and each y-coordinate by 1
∴ The image of point N (-7 , -6) is (-7 + 8 , -6 + 1) = (1 , -5)
∴ The image of point P (-4 , -3) is (-4 + 8 , - 3 + 1) = (4 , -2)
∴ The image of point Q (-4 , -6) is (-4 + 8 , -6 + 1) = (4 , -5)
* The vertices of the triangle after the translation are
(1 , -5) , (4 , -2) , (4 , -5)
- Lets find from the graph the names of these vertices
∵ Δ KLM has the same vertices k (1 , -5) , L (4 , -2) , M (4 , -5)
* The image of Δ NPQ after the translation is Δ KLM