Prove that the centroid of any triangle is located at the point of intersection of the medians. Place the axes so that the vertices are (a, 0), (0, b), and (c, 0), and so that c> a. [Hints: Recall that a median is a line segment from a vertex to the midpoint of the opposite side. Recall also that the medians intersect at a point two-thirds of the way from each vertex (along the median) to the opposite side.] Choose the position of the triangle on the axes as described above. Recall that the medians intersect at a point ✓ of the way from each vertex (along the median) to the opposite side. The median from (0, b) goes to the midpoint (a+c) ,o). So, the point of intersection of the medians is ( (a+c) of the opposite side, whose location is 2 2 This can also be verified by finding the equations of two medians and solving them simultaneously to find their point of intersection. = Find the area of the triangle. A=(c-a)b Now find the location (x, y) of the centroid of the triangle. These computations can be simplified by using horizontal rather than vertical approximating rectangles. 14° ( ((-* ус )-(- ya a+c + dy= 3 x A Jo 1 ya **** ( * ((-* + ¹)² - ( -+1)³) )*- */ dy= 3 Thus, the centroid is (x, y) = a+c b as claimed. 3
Prove that the centroid of any triangle is located at the point of intersection of the medians. Place the axes so that the vertices are (a, 0), (0, b), and (c, 0), and so that c> a. [Hints: Recall that a median is a line segment from a vertex to the midpoint of the opposite side. Recall also that the medians intersect at a point two-thirds of the way from each vertex (along the median) to the opposite side.] Choose the position of the triangle on the axes as described above. Recall that the medians intersect at a point ✓ of the way from each vertex (along the median) to the opposite side. The median from (0, b) goes to the midpoint (a+c) ,o). So, the point of intersection of the medians is ( (a+c) of the opposite side, whose location is 2 2 This can also be verified by finding the equations of two medians and solving them simultaneously to find their point of intersection. = Find the area of the triangle. A=(c-a)b Now find the location (x, y) of the centroid of the triangle. These computations can be simplified by using horizontal rather than vertical approximating rectangles. 14° ( ((-* ус )-(- ya a+c + dy= 3 x A Jo 1 ya **** ( * ((-* + ¹)² - ( -+1)³) )*- */ dy= 3 Thus, the centroid is (x, y) = a+c b as claimed. 3
Calculus: Early Transcendentals
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Author:James Stewart
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Chapter1: Functions And Models
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Hello, I am stuck on this proof problem with finding a centroid in Calculus II. I am not sure how I am suppose to set up the integral as it is wrong and not sure how to proceed.
![Prove that the centroid of any triangle is located at the point of intersection of the medians. Place the axes so that the vertices are (a, 0), (0, b), and (c, 0), and so that c>a. [Hints: Recall that a median is a line segment from a
vertex to the midpoint of the opposite side. Recall also that the medians intersect at a point two-thirds of the way from each vertex (along the median) to the opposite side.]
2
3
Choose the position of the triangle on the axes as described above. Recall that the medians intersect at a point ✓ of the way from each vertex (along the median) to the opposite side. The median from (0, b) goes to the midpoint
(a+c)
(a+c)
,0). So, the point of intersection of the medians is (
of the opposite side, whose location is
2
2
=
-40)-(1-10)
c),
This can also be verified by finding the equations of two medians and solving them simultaneously to find their point of intersection.
Find the area of the triangle.
A = (c-a)b
Now find the location (x, y) of the centroid of the triangle. These computations can be simplified by using horizontal rather than vertical approximating rectangles.
b
ус
x =
A
b
1 * ( × ( ( - * + 2 ) - ( - * + 1))
ya
a+c
dy =
b
3
b
*-*£*(
=
A
10
ус
+
b
b
3
((-* - ¦)-(-*+ ¦)')] _ )--|¦
a + c b
(x,y) - (0±0.0).
(을)
Thus, the centroid is (x,)
3
ya
b
as claimed.
×
dy](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F8b313955-89cd-4b2a-a000-54fbe962adf3%2F963b49ad-7f78-41d0-a0f2-81b4669b816d%2F0ubg29k_processed.png&w=3840&q=75)
Transcribed Image Text:Prove that the centroid of any triangle is located at the point of intersection of the medians. Place the axes so that the vertices are (a, 0), (0, b), and (c, 0), and so that c>a. [Hints: Recall that a median is a line segment from a
vertex to the midpoint of the opposite side. Recall also that the medians intersect at a point two-thirds of the way from each vertex (along the median) to the opposite side.]
2
3
Choose the position of the triangle on the axes as described above. Recall that the medians intersect at a point ✓ of the way from each vertex (along the median) to the opposite side. The median from (0, b) goes to the midpoint
(a+c)
(a+c)
,0). So, the point of intersection of the medians is (
of the opposite side, whose location is
2
2
=
-40)-(1-10)
c),
This can also be verified by finding the equations of two medians and solving them simultaneously to find their point of intersection.
Find the area of the triangle.
A = (c-a)b
Now find the location (x, y) of the centroid of the triangle. These computations can be simplified by using horizontal rather than vertical approximating rectangles.
b
ус
x =
A
b
1 * ( × ( ( - * + 2 ) - ( - * + 1))
ya
a+c
dy =
b
3
b
*-*£*(
=
A
10
ус
+
b
b
3
((-* - ¦)-(-*+ ¦)')] _ )--|¦
a + c b
(x,y) - (0±0.0).
(을)
Thus, the centroid is (x,)
3
ya
b
as claimed.
×
dy
Transcribed Image Text:O
+73-
A
b
PT OF INTERSECTION OF MEDIANS.
y=2~) (류(ㄷ), ㅎㅂ)→ (을)
(a))
C
A7
9(20)} & 99루가
13- 90 FW
9
X = Á Syll-ye+)-(2+1)) dy
A
X
4
-ys
y-0--논(x-c)
m-u-5--b
G-C
1
y-b=-=-3x
X
y-u=-317-9
노(G
y = - b (x-a) -> -3x+b
x= -ya
b
G
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