A triangle has corners at (5 ,9 )(5,9), (4 ,1 )(4,1), and (3 ,8 )(3,8). How far is the triangle's centroid from the origin?

1 Answer
Apr 23, 2018

color(blue)(2sqrt(13))213 units

Explanation:

The centroid is the point where the triangles medians meet. A median is a line through a vertex to the midpoint of the opposite side. A triangle has three medians , but will will only need to find two of these to find the point of intersection, which is the centroid.

Chose two sides:

ABAB and ACAC

Let A=(5,9) ,B=(4,1), C=(3,8)A=(5,9),B=(4,1),C=(3,8)

Find the coordinates of the midpoints of these two sides:

The coordinates of the midpoint of a line are given by:

((x_1+x_2)/2,(y_1+y_2)/2)(x1+x22,y1+y22)

For ABAB

((5+4)/2,(9+1)/2)=(9/2,5)(5+42,9+12)=(92,5)

For ACAC

((5+3)/2,(9+8)/2)=(4,17/2)(5+32,9+82)=(4,172)

ABAB passes through vertex C=(3,8)C=(3,8)

ACAC passes through vertex B=(4,1)B=(4,1)

We now find the equations of two lines using midpoints and vertices.

For ABAB

Gradient:

(8-5)/(3-9/2)=3/(-3/2)=-285392=332=2

Using point slope form of a line:

(y_2-y_1)=m(x_2-x_1)(y2y1)=m(x2x1)

y-8=-2(x-3)y8=2(x3)

y=-2x+14 \ \ \ \ \ \ \ [1]

For AC

Gradient:

(1-17/2)/(4-4)=(-15/2)/0( this is undefined and shows we have a vertical line.

x=4 \ \ \ \ \ [2]

Solving simultaneously:

y=-2(4)+14=>y=6

Coordinates of centroid:

(4,6)

To find the distance from the origin we use the distance formula:

d=sqrt((x_2-x_1)^2+(y_2-y_1)^2)

d=sqrt((0-4)^2+(0-6)^2)=sqrt(52)=2sqrt(13)

PLOT:

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