Answer:
a) The vertices of the triangle G'H'I' are [tex]G'(x,y) = (6, 1)[/tex], [tex]H'(x,y) = (6,-2)[/tex] and [tex]I'(x,y) = (1, 1)[/tex], respectively.
b) The triangle G''H''I'' have vertices at [tex]G''(x,y) = (-3,-12)[/tex], [tex]H''(x,y) = (-1, -9)[/tex] and [tex]I''(x,y) =(-6,-12)[/tex].
Step-by-step explanation:
a) From Linear Algebra, we define the translation by the following vector:
[tex]O'(x, y) = O(x,y) +T(x,y)[/tex] (1)
Where:
[tex]O(x,y)[/tex] - Original point, dimensionless.
[tex]O'(x,y)[/tex] - Translated point, dimensionless.
[tex]T(x,y)[/tex] - Translation vector, dimensionless.
If we know that [tex]G(x,y) = (-1,6)[/tex], [tex]H(x,y) =(-1,3)[/tex], [tex]I(x,y) = (-6,6)[/tex] and [tex]T(x,y) = (7,-5)[/tex], then the vertices of the triangle G'H'I' are, respectively:
[tex]G'(x,y) = G(x,y) +T(x,y)[/tex] (2)
[tex]G'(x,y) = (-1,6) +(7,-5)[/tex]
[tex]G'(x,y) = (6, 1)[/tex]
[tex]H'(x,y) = H(x,y) + T(x,y)[/tex] (3)
[tex]H'(x,y) = (-1,3) + (7,-5)[/tex]
[tex]H'(x,y) = (6,-2)[/tex]
[tex]I'(x,y) =I(x,y)+T(x,y)[/tex] (4)
[tex]I'(x,y) = (-6,6) +(7,-5)[/tex]
[tex]I'(x,y) = (1, 1)[/tex]
The vertices of the triangle G'H'I' are [tex]G'(x,y) = (6, 1)[/tex], [tex]H'(x,y) = (6,-2)[/tex] and [tex]I'(x,y) = (1, 1)[/tex], respectively.
b) From Linear Algebra, we define the reflection with respect to a horizontal line as follows:
[tex]O''(x,y) = O(x,y) - 2\cdot [O(x,y)-R(x,y)][/tex] (5)
Where:
[tex]O(x,y)[/tex] - Original point, dimensionless.
[tex]R(x,y)[/tex] - Reflection point, dimensionless.
If we know that [tex]G(x,y) = (-1, 6)[/tex] and [tex]R(x,y) = (-1,-3)[/tex], then the location of point G'' is:
[tex]G''(x,y) =G(x,y) -2\cdot [G(x,y)-R(x,y)][/tex] (6)
[tex]G''(x,y) = (-1,6)-2\cdot [(-1,6)-(-1,-3)][/tex]
[tex]G''(x,y) = (-1,6) -2\cdot (0,9)[/tex]
[tex]G''(x,y) = (-1,6)-(2,18)[/tex]
[tex]G''(x,y) = (-3,-12)[/tex]
If we know that [tex]H(x,y) =(-1,3)[/tex] and [tex]R(x,y) = (-1,-3)[/tex] , then the location of the point H'' is:
[tex]H''(x,y) =H(x,y) -2\cdot [H(x,y)-R(x,y)][/tex] (7)
[tex]H''(x,y) = (-1,3)-2\cdot [(-1,3)-(-1,-3)][/tex]
[tex]H''(x,y) = (-1, 3) -2\cdot (0,6)[/tex]
[tex]H''(x,y) = (-1,3)-(0,12)[/tex]
[tex]H''(x,y) = (-1, -9)[/tex]
If we know that [tex]I(x,y) = (-6,6)[/tex] and [tex]R(x,y) = (-6, -3)[/tex], then the location of the point I'' is:
[tex]I''(x,y) =I(x,y) -2\cdot [I(x,y)-R(x,y)][/tex] (8)
[tex]I''(x,y) = (-6,6)-2\cdot [(-6,6)-(-6,-3)][/tex]
[tex]I''(x,y) = (-6,6)-2\cdot (0,9)[/tex]
[tex]I''(x,y) = (-6,6)-(0,18)[/tex]
[tex]I''(x,y) =(-6,-12)[/tex]
The triangle G''H''I'' have vertices at [tex]G''(x,y) = (-3,-12)[/tex], [tex]H''(x,y) = (-1, -9)[/tex] and [tex]I''(x,y) =(-6,-12)[/tex].
