Inactive Tutor answered • 05/24/15
Tutor
New to Wyzant
Can you picture these two rods as two of the sides of a right triangle? The third side would be formed by connecting the two ends that are not at the origin. This question is asking where (the ordered pair) the centroid of that triangle would be. Since the Centroid is formed at the concurrence of the medians, try drawing this triangle on a piece of graph paper; remember that the problem tells you that the legs of the right triangle are of equal length. Then connect each vertex to the center of its opposite side. When you do this, you should see that the crossing point (the centroid) is closer to the origin than given in Jose's answer above.
To actually solve for this value, I first found the slopes of the median lines, using
m = (y2 - y1)/(x1 - x2).
One line goes from the x-axis @ (L,0) across to the y axis @ (0, L/2)
This line has a slope of -1/2.
The other median goes from the x axis @ (L/2, 0) to the y axis at (0,L)
this line has a slope of -2.
Then I made the two equations of these lines in y=mx + b form.
one is y = -1/2x + L/2 and the other is y = -2x + L
Then I solved for the x value where these two cross by setting those two equations equal to each other (Because the ordered pairs at an intersection of two crossing lines are equal!).
-2x + L = (-1/2)x + (L/2)
multiply through everything by 2 to eliminate the fraction:
-4x + 2L = -x + L
Solve for x:
I get x = L/3
Since we started with a triangle whose two perpendicular sides are equal.... go back to your graph and look at it.... I know that the y value is the same... y = L/3.
So, the ordered pair of the centroid is (L/3, L/3)
