Question

Enter the ordered pair for the vertices for T(3, –2)(QRST).

Answer 1

Given:

The figure QRST.

The rule of translation is $T_{(3,-2)}(QRST)$.

To find:

The coordinate of vertices after translation.

Solution:

According to the given rule of translation is $T_{(3,-2)}(QRST)$, we get

$(x,y)\to (x+3,y-2)$

From the given figure it is clear that the vertices of QRST are Q(1,3), R(3,-3), S(0,-2) and T(-2,1).

Using the above rule, we get

$Q(1,3)\to Q'(1+3,3-2)=Q'(4,1)$

$R(3,-3)\to R'(3+3,-3-2)=R'(6,-5)$

$S(0,-2)\to S'(0+3,-2-2)=S'(3,-4)$

$T(-2,1)\to T'(-2+3,1-2)=T'(1,-1)$

Therefore, the coordinates of the vertices after translation are Q'(4,1), R'(6,-5), S'(3,-4) and T'(1,-1).

Answer 2

The ordered pairs for the vertices of T(3, –2)(QRST) are

$\mathbf{Q' = (4,1)}$  $\mathbf{R' = (6,-5)}$  $\mathbf{S' = (3,-4)}$ and $\mathbf{T' = (1, -1)}$

The coordinates of QRST are:

$\mathbf{Q = (1,3)}$

$\mathbf{R = (3,-3)}$

$\mathbf{S = (0,-2)}$

$\mathbf{T = (-2,1)}$

The transformation rule of $\mathbf{T(3,-2)}$ is:

$\mathbf{(x,y) \to (x + 3,y -2)}$

So, we have:

$\mathbf{Q' = (1 + 3,3 -2)}$

$\mathbf{Q' = (4,1)}$

$\mathbf{R' = (3 + 3,-3 -2)}$

$\mathbf{R' = (6,-5)}$

$\mathbf{S' = (0 + 3,-2 -2)}$

$\mathbf{S' = (3,-4)}$

$\mathbf{T' = (-2 + 3,1 -2)}$

$\mathbf{T' = (1, -1)}$

Hence, the ordered pairs for the vertices of T(3, –2)(QRST) are

$\mathbf{Q' = (4,1)}$  $\mathbf{R' = (6,-5)}$  $\mathbf{S' = (3,-4)}$ and $\mathbf{T' = (1, -1)}$

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