Finding the unknown points of a second triangle

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Triangle A has vertices at points (2,3), (0,0), and (2,0) with a center calculated as (2/3,1). The task is to find the vertices of Triangle B, which has a center at (4/3,3). The vector between the centers of the two triangles is determined to be <2/3, 2>, and its magnitude is calculated as (2*√10)/3. The solution involves adding this magnitude to the points of Triangle A, but the user struggles to achieve the correct x-coordinate for the center of Triangle B with the new points. The relationship between Triangle A and Triangle B remains unclear.
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Homework Statement


Triangle A has three points a(2,3)b(0,0)c(2,0) and its center is (2/3,1). Find the other three points of Triangle B with a center of (4/3,3).


Homework Equations


Center of a triangle:
x = ax+bx+cx /3
y = ay+by+cy / 3

Magnitude = <a,b> , √(a^2+b^2)

The Attempt at a Solution


1)Obtaining the vector of AB by subtracting the two center <4/3-2/3, 3-1> = <2/3,2>
2)Magnitude of vector AB is (2*√10)/3
3)Adding the magnitude to the three points of triangle A
4)Check answer by calculating the center with the 3 new points, should match with the given center above.

Step 4 is where I'm stuck, I can't get the same x center.
 
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Will15 said:

Homework Statement


Triangle A has three points a(2,3)b(0,0)c(2,0) and its center is (2/3,1). Find the other three points of Triangle B with a center of (4/3,3).


Homework Equations


Center of a triangle:
x = ax+bx+cx /3
y = ay+by+cy / 3

Magnitude = <a,b> , √(a^2+b^2)

The Attempt at a Solution


1)Obtaining the vector of AB by subtracting the two center <4/3-2/3, 3-1> = <2/3,2>
2)Magnitude of vector AB is (2*√10)/3
3)Adding the magnitude to the three points of triangle A
4)Check answer by calculating the center with the 3 new points, should match with the given center above.

Step 4 is where I'm stuck, I can't get the same x center.

How is triangle B related to triangle A?
 
Mark44 said:
How is triangle B related to triangle A?

The relation is unknown.
 
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