contestada

Triangle QRS has vertices Q(-4,2),R(3,0), and S(4,3).
If AQRSis translated 4 units down and 6 units right to
create 스Q'R'S', what are the coordinates of the
vertices of AQ'R'S'?

Respuesta :

xero099

The coordinates of the vertices of [tex]\triangle Q'R'S'[/tex] are [tex]Q'(x,y) = (0, -4)[/tex], [tex]R'(x,y) = (7,-6)[/tex] and [tex]S'(x,y) = (8, -3)[/tex], respectively.

Vectorially speaking, a point is translated to another location on Cartesian plane by means of this expression:

[tex]A'(x,y) = A(x,y) + T(x,y)[/tex] (1)

Where:

  • [tex]A(x,y)[/tex] - Original point.
  • [tex]A'(x,y)[/tex] - Translated point.
  • [tex]T(x,y)[/tex] - Translation vector.

If we know that [tex]Q(x,y) = (-4,2)[/tex], [tex]R(x,y) = (3,0)[/tex], [tex]S(x,y) = (4, 3)[/tex] and [tex]R'(x,y) = (7, -6)[/tex][tex]T(x,y) = (4, -6)[/tex], then the coordinates of the vertices of the triangle [tex]Q'R'S'[/tex] are:

[tex]Q'(x,y) = Q(x,y) + T(x,y)[/tex]

[tex]Q'(x,y) = (-4,2) + (4,-6)[/tex]

[tex]Q'(x,y) = (0, -4)[/tex]

[tex]R'(x,y) = R(x,y) + T(x,y)[/tex]

[tex]R'(x,y) = (3,0) +(4,-6)[/tex]

[tex]R'(x,y) = (7,-6)[/tex]

[tex]S'(x,y) = S(x,y) + T(x,y)[/tex]

[tex]S'(x,y) = (4, 3) + (4, -6)[/tex]

[tex]S'(x,y) = (8, -3)[/tex]

The coordinates of the vertices of [tex]\triangle Q'R'S'[/tex] are [tex]Q'(x,y) = (0, -4)[/tex], [tex]R'(x,y) = (7,-6)[/tex] and [tex]S'(x,y) = (8, -3)[/tex], respectively.

We kindly invite to see this question on translations: https://brainly.com/question/17485121