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igomit [66]
3 years ago
12

Given a triangle ABC at A = ( - 1, 3 ) B = ( 1, - 1 ) C = ( 2, 2 ) and if the triangle is rotated 90 ° clockwise about the point

( 4, 1 ), find the new location of the point A'.

Mathematics
1 answer:
lidiya [134]3 years ago
7 0

Answer:

The coordinates of A' are (6,6).

Step-by-step explanation:

It is given that in triangle ABC, A = (-1, 3), B = (1, - 1) and C = (2, 2).

If a point is rotated 90 ° clockwise about the point ( a,b ), then

(x,y)\rightarrow ((y-b)+a,-(x-a)+b)

It is given that the triangle is rotated 90 ° clockwise about the point ( 4, 1 ). So, a=4 and b=1.

(x,y)\rightarrow ((y-1)+4,-(x-4)+1)

(x,y)\rightarrow (y+3,-x+5)

The coordinates of A are (-1,3), So, the coordinates of A' are

A(-1,3)\rightarrow A'(3+3,-(-1)+5)

A(-1,3)\rightarrow A'(6,6)

Therefore the coordinates of A' are (6,6).

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Answer:

a_n=-3(3)^{n-1} ; {-3,-9, -27,- 81, -243, ...}

a_n=-3(-3)^{n-1} ; {-3, 9,-27, 81, -243, ...}

a_n=3(\frac{1}{2})^{n-1} ; {3, 1.5, 0.75, 0.375, 0.1875, ...}

a_n=243(\frac{1}{3})^{n-1} ; {243, 81, 27, 9, 3, ...}

Step-by-step explanation:

The first explicit equation is

a_n=-3(3)^{n-1}

At n=1,

a_1=-3(3)^{1-1}=-3

At n=2,

a_2=-3(3)^{2-1}=-9

At n=3,

a_3=-3(3)^{3-1}=-27

Therefore, the geometric sequence is {-3,-9, -27,- 81, -243, ...}.

The second explicit equation is

a_n=-3(-3)^{n-1}

At n=1,

a_1=-3(-3)^{1-1}=-3

At n=2,

a_2=-3(-3)^{2-1}=9

At n=3,

a_3=-3(-3)^{3-1}=-27

Therefore, the geometric sequence is {-3, 9,-27, 81, -243, ...}.

The third explicit equation is

a_n=3(\frac{1}{2})^{n-1}

At n=1,

a_1=3(\frac{1}{2})^{1-1}=3

At n=2,

a_2=3(\frac{1}{2})^{2-1}=1.5

At n=3,

a_3=3(\frac{1}{2})^{3-1}=0.75

Therefore, the geometric sequence is {3, 1.5, 0.75, 0.375, 0.1875, ...}.

The fourth explicit equation is

a_n=243(\frac{1}{3})^{n-1}

At n=1,

a_1=243(\frac{1}{3})^{1-1}=243

At n=2,

a_2=243(\frac{1}{3})^{2-1}=81

At n=3,

a_3=243(\frac{1}{3})^{3-1}=27

Therefore, the geometric sequence is {243, 81, 27, 9, 3, ...}.

6 0
3 years ago
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