Answer:
a) The vertices of the triangle G'H'I' are
,
and
, respectively.
b) The triangle G''H''I'' have vertices at
,
and
.
Step-by-step explanation:
a) From Linear Algebra, we define the translation by the following vector:
(1)
Where:
- Original point, dimensionless.
- Translated point, dimensionless.
- Translation vector, dimensionless.
If we know that
,
,
and
, then the vertices of the triangle G'H'I' are, respectively:
(2)
![G'(x,y) = (-1,6) +(7,-5)](https://tex.z-dn.net/?f=G%27%28x%2Cy%29%20%3D%20%28-1%2C6%29%20%2B%287%2C-5%29)
![G'(x,y) = (6, 1)](https://tex.z-dn.net/?f=G%27%28x%2Cy%29%20%3D%20%286%2C%201%29)
(3)
![H'(x,y) = (-1,3) + (7,-5)](https://tex.z-dn.net/?f=H%27%28x%2Cy%29%20%3D%20%28-1%2C3%29%20%2B%20%287%2C-5%29)
![H'(x,y) = (6,-2)](https://tex.z-dn.net/?f=H%27%28x%2Cy%29%20%3D%20%286%2C-2%29)
(4)
![I'(x,y) = (-6,6) +(7,-5)](https://tex.z-dn.net/?f=I%27%28x%2Cy%29%20%3D%20%28-6%2C6%29%20%2B%287%2C-5%29)
![I'(x,y) = (1, 1)](https://tex.z-dn.net/?f=I%27%28x%2Cy%29%20%3D%20%281%2C%201%29)
The vertices of the triangle G'H'I' are
,
and
, respectively.
b) From Linear Algebra, we define the reflection with respect to a horizontal line as follows:
(5)
Where:
- Original point, dimensionless.
- Reflection point, dimensionless.
If we know that
and
, then the location of point G'' is:
(6)
![G''(x,y) = (-1,6)-2\cdot [(-1,6)-(-1,-3)]](https://tex.z-dn.net/?f=G%27%27%28x%2Cy%29%20%3D%20%28-1%2C6%29-2%5Ccdot%20%5B%28-1%2C6%29-%28-1%2C-3%29%5D)
![G''(x,y) = (-1,6) -2\cdot (0,9)](https://tex.z-dn.net/?f=G%27%27%28x%2Cy%29%20%3D%20%28-1%2C6%29%20-2%5Ccdot%20%280%2C9%29)
![G''(x,y) = (-1,6)-(2,18)](https://tex.z-dn.net/?f=G%27%27%28x%2Cy%29%20%3D%20%28-1%2C6%29-%282%2C18%29)
![G''(x,y) = (-3,-12)](https://tex.z-dn.net/?f=G%27%27%28x%2Cy%29%20%3D%20%28-3%2C-12%29)
If we know that
and
, then the location of the point H'' is:
(7)
![H''(x,y) = (-1, 3) -2\cdot (0,6)](https://tex.z-dn.net/?f=H%27%27%28x%2Cy%29%20%3D%20%28-1%2C%203%29%20-2%5Ccdot%20%280%2C6%29)
![H''(x,y) = (-1,3)-(0,12)](https://tex.z-dn.net/?f=H%27%27%28x%2Cy%29%20%3D%20%28-1%2C3%29-%280%2C12%29)
![H''(x,y) = (-1, -9)](https://tex.z-dn.net/?f=H%27%27%28x%2Cy%29%20%3D%20%28-1%2C%20-9%29)
If we know that
and
, then the location of the point I'' is:
(8)
![I''(x,y) = (-6,6)-2\cdot [(-6,6)-(-6,-3)]](https://tex.z-dn.net/?f=I%27%27%28x%2Cy%29%20%3D%20%28-6%2C6%29-2%5Ccdot%20%5B%28-6%2C6%29-%28-6%2C-3%29%5D)
![I''(x,y) = (-6,6)-2\cdot (0,9)](https://tex.z-dn.net/?f=I%27%27%28x%2Cy%29%20%3D%20%28-6%2C6%29-2%5Ccdot%20%280%2C9%29)
![I''(x,y) = (-6,6)-(0,18)](https://tex.z-dn.net/?f=I%27%27%28x%2Cy%29%20%3D%20%28-6%2C6%29-%280%2C18%29)
![I''(x,y) =(-6,-12)](https://tex.z-dn.net/?f=I%27%27%28x%2Cy%29%20%3D%28-6%2C-12%29)
The triangle G''H''I'' have vertices at
,
and
.