Let be a triangle with . Suppose is the midpoint of . The distance of the centroid of from the point of intersection of the line and is:
- A
- B
- C
- D
Let be a triangle with . Suppose is the midpoint of . The distance of the centroid of from the point of intersection of the line and is:
Correct answer:C
Standard Method
Given: and is the midpoint of .
Find: The distance between the centroid of and the point of intersection of the two given lines.
Use the fact that the centroid of a triangle lies on the median joining a vertex to the midpoint of the opposite side and divides that median in the ratio from the vertex. Thus, if is the midpoint of , then the centroid lies on and divides it internally in the ratio .
So,
Now find the intersection point of the lines.
For the first line,
For the second line,
At the point of intersection, coordinates are equal. From ,
Then from the second line,
Hence the intersection point is
Now use the distance formula between and :
Therefore, the distance is and the correct option is C.
Median-Centroid Shortcut
Given: is the midpoint of and is the third vertex.
Find: The required distance.
Instead of introducing coordinates of and , directly use the centroid property on median . The centroid is the section point that divides internally in the ratio from toward .
Thus,
The line intersection is obtained quickly by noticing that the first line has always. On the second line, set , giving . Hence the intersection point is
Now compute
Therefore,
Hence the correct option is C.
Treating the centroid as the average of and is incorrect because is not a vertex of the triangle. Use the fact that the centroid divides the median in the ratio from the vertex .
Misreading can lead to invalid parameterization. It means for the entire first line, while the direction vector is .
Using the wrong section ratio for the centroid is a common error. The centroid divides a median in the ratio from the vertex, not from the vertex.
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