what are the coordinates of the centroid of each triangle given the three vertices of D(0,3),C(-1,0),B(4,0)?
The quickest, but little know method is to simply "average" the x's and the y's
x for centroid = (0-1+4)/3 = 1
y for centroid = (3+0+0)/3 = 1
the centroid is (1,1)
Slightly more complicated way:
find midpoint of DB = M(2 , 3/2)
join CM
Let the centroid be P(a,b)
we know that P must be such that CP : PM = 2:1
(a+1)/2 = (2-a)/1
a+1 = 4 - 2a
3a = 3
a = 1
(b-0)/2 = (3/2 - b)/1
b = 3 - 2b
3b=3
b = 1
centroid = P(a,b) = P(1,1)
most complicated way:
find the equation for CM
after finding the midpoint N of BC, find the equation for DN
intersect DN and CM
I will not do it, but am absolutely sure it will be (1,1)
To find the coordinates of the centroid of a triangle, you can use the following formula:
Centroid(X, Y) = ((X1 + X2 + X3)/3, (Y1 + Y2 + Y3)/3)
Where (X1, Y1), (X2, Y2), and (X3, Y3) are the coordinates of the vertices of the triangle.
Let's apply this formula to find the centroid coordinates of the triangle with vertices D(0,3), C(-1,0), and B(4,0):
X-coordinate of the centroid:
X = (0 + (-1) + 4)/3
= 3/3
= 1
Y-coordinate of the centroid:
Y = (3 + 0 + 0)/3
= 3/3
= 1
Therefore, the coordinates of the centroid of the triangle DCB are (1, 1).