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wariber [46]
3 years ago
10

TOY has coordinates T (-3, 4), O (-4, 1) and Y (-2, 3). a translation maps point T to T' (-1, 1). find the coordinates of O' and

Y' under this translation.​
Mathematics
1 answer:
NeX [460]3 years ago
8 0

Answer:

Answer:  The correct option is (A) O' (−2, −2); Y' (0, 0).

Step-by-step explanation:  Given that the co-ordinates of the vertices of ΔTOY are T(−3, 4), O (−4, 1), and Y (−2, 3). A translation maps point T to T' (−1, 1).

We are to find the co-ordinates of the points O' and Y'.

The given transformation from T to T' is

T(−3, 4)   ⇒   T' (−1, 1).

Let,  (−3 + x, 4 + y) =  (-1, 1).

So,

and

That is, the transformation rule is

(a, b) ⇒ (a+2, b-3).

Therefore,

co-ordinates of O' are (-4+2, 1-3) = (-2, -2),

and

co-ordinates of Y' are (-2+2, 3-3) = (0, 0).

Thus, the required co-ordinates of O' and Y' are (-2, -2) and (0, 0) respectively.

Option (A) is correct.

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P=\left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right]

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Transition matrix P raised to the power 2 (at k = 2)

P^{2} =\left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right] X \left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right]

P^{2} =\left[\begin{array}{ccc}0.2203&0.7797\\0.2034&0.7966\end{array}\right]

Transition matrix P raised to the power 3 (at k = 3)

P^{3} =\left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right] X \left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right]X\left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right]

P^{3} =\left[\begin{array}{ccc}0.2203&0.7797\\0.2034&0.7966\end{array}\right] X \left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right]

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Transition matrix P raised to the power 4 (at k = 4)

P^{4} =\left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right] X \left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right]X\left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right]X\left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right]

P^{4} =\left[\begin{array}{ccc}0.2086&0.7914\\0.2064&0.7936\end{array}\right] X \left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right]

P^{4} =\left[\begin{array}{ccc}0.2071&0.7929\\0.2068&0.7932\end{array}\right]

Transition matrix P raised to the power 5 (at k = 5)

P^{5} =\left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right] X \left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right]X\left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right]X\left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right]X\left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right]

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P^{5} =\left[\begin{array}{ccc}0.2069&0.7931\\0.2069&0.7931\end{array}\right]

P⁵ at k = 5 both the rows identical. Hence the stationary matrix S is:

S = [ 0.2069 , 0.7931 ]

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