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# Find the triangle's centre of rotation....?

The vertices of a triangle ABC have the coordinates

A (0, 0), B (3, 0) and C (0, 4)

Under a rotation the triangle is mapped onto the triangle A'B'C' whose vertices have the coordinates

A' (9, 0), B' (9, 3) and C' (5, 0)

Find the centre of rotation R? Prove that angle ARA' is 90 degrees and the y-coordinate of R is 4.5.

Thanks.

### 2 Answers

- 12 months agoFavorite Answer
You have some point (h , k)

The distance from (h , k) to A and A' is a

The distance from (h , k) to B and B' is b

The distance from (h , k) to C and C' is c

(x - h)^2 + (y - k)^2 = r^2

(0 - h)^2 + (0 - k)^2 = a^2 = (9 - h)^2 + (9 - k)^2

(3 - h)^2 + (0 - k)^2 = b^2 = (9 - h)^2 + (3 - k)^2

(0 - h)^2 + (4 - k)^2 = c^2 = (5 - h)^2 + (0 - k)^2

h^2 + k^2 = (9 - h)^2 + (9 - k)^2

(3 - h)^2 + k^2 = (9 - h)^2 + (3 - k)^2

h^2 + (4 - k)^2 = (5 - h)^2 + k^2

h^2 + k^2 - ((3 - h)^2 + k^2) = (9 - h)^2 + (9 - k)^2 - (9 - h)^2 - (3 - k)^2

h^2 - (3 - h)^2 + k^2 - k^2 = (9 - k)^2 - (3 - k)^2

h^2 - (9 - 6h + h^2) = 81 - 18k + k^2 - (9 - 6k + k^2)

h^2 - 9 + 6h - h^2 = 81 - 9 - 18k + 6k + k^2 - k^2

6h - 9 = 72 - 12k

6h + 12k = 81

2h + 4k = 27

2h = 27 - 4k

h = (27 - 4k) / 2

h^2 + k^2 = (9 - h)^2 + (9 - k)^2

h^2 + (4 - k)^2 = (5 - h)^2 + k^2

h^2 + k^2 - h^2 - (4 - k)^2 = (9 - h)^2 + (9 - k)^2 - (5 - h)^2 - k^2

k^2 - (16 - 8k + k^2) = 81 - 18h + h^2 + 81 - 18k + k^2 - (25 - 10h + h^2) - k^2

k^2 - k^2 + 8k - 16 = 81 + 81 - 25 - 18h - 18k + 10h + h^2 - h^2 + k^2 - k^2

8k - 16 = 162 - 25 - 8h - 18k

18k + 8k + 8h = 162 + 16 - 25

26k + 8h = 178 - 25

26k + 8h = 153

h = (27 - 4k) / 2

26k + 8 * (1/2) * (27 - 4k) = 153

26k + 4 * (27 - 4k) = 153

26k - 16k + 108 = 153

10k = 45

k = 4.5

h = (27 - 4k) / 2

h = (27 - 18) / 2

h = 9/2

h = 4.5

(4.5 , 4.5) is the center of rotation.

h^2 + k^2 = a^2

4.5^2 + 4.5^2 = a^2

2 * 4.5^2 = a^2

4.5 * sqrt(2) = a

We know that the distance between A and A' is 9 units

9^2 = (4.5 * sqrt(2))^2 + (4.5 * sqrt(2))^2 - 2 * 4.5 * sqrt(2) * 4.5 * sqrt(2) * cos(T)

81 = 4.5^2 * 2 + 4.5^2 * 2 - 2 * 2 * 4.5^2 * cos(T)

81 = 4 * 4.5^2 - 4 * 4.5^2 * cos(T)

81 = 4 * 20.25 - 4 * 20.25 * cos(T)

81 = 81 - 81 * cos(T)

0 = -81 * cos(T)

0 = cos(T)

T = cos-1(0)

T = 90 degrees

- atsuoLv 612 months ago
Please draw a diagram on the xy-plane .

Let the center of rotation be R(p,q) .

A(0,0) is moved to A'(9,0) by the rotation , so RA = RA' . That is ,

R(p,q) exists on the perpendicular bisector PB1 of AA' .

PB1 passes through the midpoint of AA' , it is (9/2,0) .

And PB1 is vertical , so p = 9/2 .

C(0,4) is moved to C'(5,0) by the rotation , so RC = RC' . That is ,

R(9/2,q) exists on the perpendicular bisector PB2 of CC' .

PB2 passes through the midpoint of CC' , it is (5/2,2) .

And the slope of CC' is -4/5 , so the slope of PB2 is 5/4 .

Therefore the equation of PB2 is

(y - 2) = (5/4)(x - 5/2)

y = (5/4)x - 9/8

R(9/2,q) is on PB2 , so q = (5/4)(9/2) - 9/8 = 9/2 .

This is the y-coordinate of R .

Next , think about the angle ARA' .

The slope of AR is 45° , and the slope of RA' is -45° .

Therefore , the angle ARA' is 90° .