contestada

Triangle PQR is translated to produce Triangle P'Q'R', as shown. Using the same
translation, Square STUV will be translated to produce Square S'T'U'V'.

What will be the coordinates of Point T'?
(-6,-7)
(-5,-6)
(-5,-2)
(-1,-6)

Triangle PQR is translated to produce Triangle PQR as shown Using the same translation Square STUV will be translated to produce Square STUV What will be the co class=

Respuesta :

The coordinates of the point  are (-1, -6).

Step-by-step explanation:

Step 1:

We need to plot the points for both triangles PQR and [tex]P^{1} Q^{1} R^{1}[/tex].

The points of the triangle PQR are P (1, 6), Q (3, 2), and R (1, 2).

The points of the triangle [tex]P^{1} Q^{1} R^{1}[/tex] are [tex]P^{1}[/tex] (-6, 7), [tex]Q^{1}[/tex](-4, 3), and [tex]R^{1}[/tex](-6, 3).

Step 2:

Now we determine translation of the coordinates.

P (1, 6) becomes [tex]P^{1}[/tex] (-6, 7).

Q (3, 2) becomes [tex]Q^{1}[/tex](-4, 3).

R (1, 2) becomes [tex]R^{1}[/tex](-6, 3).

So the translation is (x, y) becomes (x - 7, y + 1).

Step 3:

Now we determine the coordinates of square STUV.

The points of the square STUV are S (6, -3), T (6, -7), U (2, -3), and         V (2, -7).

The translation is (x, y) becomes (x - 7, y + 1).

For S (6, -3). x = 6 - 7 = -1, y = -3 + 1 = -2, So [tex]S^{1}[/tex] becomes (-1, -2).

For T (6, -7). x = 6 - 7 = -1, y = -7 + 1 = -6, So [tex]T^{1}[/tex] becomes  (-1, -6).

For U (2, -7). x = 2 - 7 = -5, y = -7 + 1 = -6, So [tex]U^{1}[/tex] becomes (-5, -6).

For V (2, -3). x = 2 - 7 = -5, y = -3 + 1 = -2, So [tex]V^{1}[/tex] becomes (-5, -2).

So coordinates of the point [tex]T^{1}[/tex] are (-1, -6).